RD Sharma Class 12 Ex 33.1 Solutions Chapter Chapter 33 Binomial Distribution

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter33
Exercise33.1
CategoryRD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 33.1 Solutions Chapter Chapter 33 Binomial Distribution

Question 1. There are 6% defective items in a large bulk of items. Find the probability that a sample of 8 items will include not more than one defective item.

Solution:

Let us consider X be the number of defective items in a sample of 8 items. 

So, a binomial distribution follows by x with n = 8.

and Probability of getting a defective item(p) = 0.06 

The probability of getting a defective item (1 – p) = 0.94

So, P(X = r) = 8C(0.06)(0 . 94 )8 – r, where r = 0, 1, 2, 3, . . . 8

So, the required probability is

= P (X < 1) 

= P(X = 0) + P(X = 1)

8C0 (0.06)0 (0.94)8-0 + 8C1 (0.06)1 (0.94 )8-1

= (0.94)8 + 8 (0.06) (0.94)

= (0.94)7 {0.94 + 0.48}

= 1.42 (0.94)7

Question 2. A coin is tossed 5 times. What is the probability of getting at least 3 heads?

Solution:

Let us consider X be the number of heads in 5 tosses.

So, a binomial distribution follows by x with n = 5

Probability of getting a head(p) = 1/2

And q = 1 – p = 1 – 1/2 = 1/2

P(X = r) = 5Cr (1/2)r (1/2)5-r , where r = 0, 1, 2 . . . 5

So, the required probability is

= P(X > 3)

= P(X = 3) + P(X = 4) + P(X = 5)

5C(1/2)3 (1/2)5-3 5C4 (1/2)1 (1/2)5-1 5C5 (1/2)5 (1/2)5-0

= 10 (1/2)5 + 5 (1/2)5 + 1 (1/2)5

= (1/2)5 (10 + 5 + 1)

= (1/2)5 (16)

= 1/2

Question 3. A coin is tossed 5 times. What is the probability that tail appears an odd number of times?

Solution:

Let us consider X be the number of tails when a coin is tossed 5 times.

So, a binomial distribution follows by x with n  = 5

Probability of getting head(p) = 1/2.

Also, q = 1 – p = 1/2

P(X = r) = 5C3 (1/2)r (1/2)n-r = 5Cr (1/2)5

So, the required probability is

= P(X = 1) + P(X = 3) + P(X = 5)

5C1 (1/2)5 + 5C3 (1/2)5 + 5C5 (1/2)5

= (1/2)5 [5 + 10 + 1]

= 16/32

= 1/2

Question 4. A pair of dice is thrown 6 times. If getting a total of 9 is considered a success, what is the probability of at least 5 successes?

Solution:

Let us consider X be the number of successes in 6 throws of the two dice.

So the probability of success is equal to the probability of getting a total of 9  

= Probability of getting (3, 6), (4, 5), (5, 4), (6, 3) from 36 outcomes

Here, p = 4/36 = 1/9

Also q = 1 – p = 8/9 and n = 6

Now, a binomial distribution X follows with n = 6, p = 1/9, and q = 8/9

P(X = r) = 6Cr (1/9)r (8/9)6-r

The required probability is

= P(X > 5)

= P(X = 5) + P(X = 6)

^{6}{}{C}_5 \left( \frac{1}{9} \right)^5 \left( \frac{8}{9} \right)^{6 - 5} + ^{6}{}{C}_6 \left( \frac{1}{9} \right)^6 \left( \frac{8}{9} \right)^{6 - 6}

\frac{6(8) + 1}{9^6}

= 49/96

Question 5. A fair coin is tossed 8 times, find the probability: 

(i) of exactly 5 heads.

Solution:

Let us consider X be the number of heads obtained when a fair is tossed 8 times.

Now, a binomial distribution X follows with n  = 8.

Here, p = 1/2 and q = 1 – 1/2 = 1/2

So, P(X = r) = ^8 C_r \left( \frac{1}{2} \right)^{8 - r} \left( \frac{1}{2} \right)^r

8Cr (1/2)8 , where r = 0, 1, 2, . . . , 8

So, the probability of obtaining exactly 5 heads is 

P(X = 5) = 8C5 (1/2)8

= 56/256

= 7/32

(ii) of at least six heads.

Solution:

Let us consider X be the number of heads obtained when a fair is tossed 8 times.

Now, a binomial distribution X follows with n = 8.

Here, p = 1/2 and q = 1 – 1/2 = 1/2

So, P(X = r) = ^8 C_r \left( \frac{1}{2} \right)^{8 - r} \left( \frac{1}{2} \right)^r

8Cr (1/2)8 , where r = 0, 1, 2, . . . , 8

So, the probability of obtaining exactly 5 heads is 

P(X = 5) = 8C5 (1/2)8

= 56/256

= 7/32

(iii) of at most six heads.

Solution:

Let us consider X denotes the number of heads obtained when a fair is tossed 8 times.

Now, a binomial distribution X follows with n = 8.

Here p = 1/2 and q = 1 – 1/2 = 1/2

P(X = r) = ^8 C_r \left( \frac{1}{2} \right)^{8 - r} \left( \frac{1}{2} \right)^r

8Cr (1/2)8 , r = 0, 1, 2, . . . , 8

so the probability of obtaining at most 6 heads is

P (X < 6) = 1 – [P(X = 7) + P(X = 8)]

1 - \left[ {}^8 C_7 \left( \frac{1}{2} \right)^8 +^8 C_8 \left( \frac{1}{2} \right)^8 \right]

= 1 – (8/256 + 1/256)

= 1 – 9/256

= 247/256

Question 6. Find the probability of 4 turning up at least once in two tosses of a fair die.

Solution:

Let us consider X denotes the probability of getting 4 in two tosses of a fair die.

Now, a binomial distribution X follows with n = 2.

Here, p = 1/6 and q = 5/6

P(X = r) = ^{2}{}{C}_r \left( \frac{1}{6} \right)^r \left( \frac{5}{6} \right)^{2 - r}

So, the probability of obtaining 4 at least once is

P(X > 1) = 1 – P(X = 0)

= 1 – ^{2}{}{C}_0 \left( \frac{1}{6} \right)^0 \left( \frac{5}{6} \right)^{2 - 0}

= 1 – 25/36

= 11/36

Question 7. A coin is tossed 5 times. What is the probability that head appears an even number of times?

Solution:

Let us consider X denotes the number of heads that appear when a coin is tossed 5 times.

Now, a binomial distribution X follows with n = 5 

Here p = q = 1/2.

P(X = r) = ^{5}{}{C}_r \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^{5 - r}

5Cr (1/2)5

P (head appears an even number of  times) = P(X = 0) + P(X = 2) + P(X = 4)

^{5}{}{C}_0 \left( \frac{1}{2} \right)^5 + ^{5}{}{C}_2 \left( \frac{1}{2} \right)^5 + ^{5}{}{C}_4 \left( \frac{1}{2} \right)^5

\frac{1 + 10 + 5}{2^5}

= 16/32

= 1/2

Question 8. The probability of a man hitting a target is 1/4. If he fires 7 times, what is the probability of his hitting the target at least twice?

Solution:

Let us consider X denotes the number of times the target is hit. 

Now, a binomial distribution X follows with n = 7.

Here, p = 1/4 and q = 3/4.

P(X = r) = ^{7}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^{7 - r}

P(hit the target at least twice) = P(X > 2) 

= 1 – {P(X = 0) + P(X = 1)}

1 -^{7}{}{C}_0 \left( \frac{1}{4} \right)^0 \left( \frac{3}{4} \right)^{7 - 0} - ^{7}{}{C}_1 \left( \frac{1}{4} \right)^1 \left( \frac{3}{4} \right)^{7 - 1}

= 1 – (3/4)7 – 7 (1/4) (3/4)6

= 1 – 1/16384(2187 + 5103)

= 1 – 3645/8192

= 4547/8192

Question 9. Assume that on average one telephone number out of 15 called between 2 P.M. and 3 P.M. on weekdays is busy. What is the probability that if six randomly selected telephone numbers are called, at least 3 of them will be busy?

Solution:

Let us consider X denotes the number of busy calls for 6 randomly selected telephone numbers.

Now, a binomial distribution X follows with n = 6.

Here, p = one out of 15 = 1/15

And q = 1 – 1/15 = 14/15

P(X = r) = ^{6}{}{C}_r \left( \frac{1}{15} \right)^r \left( \frac{14}{15} \right)^{6 - r}

So, the probability that at least 3 of them are busy is 

P(X > 3) = 1 – {P(X = 0) + P(X = 1) + P(X = 2)}

1 - \left\{ ^{6}{}{C}_0 \left( \frac{1}{15} \right)^0 \left( \frac{14}{15} \right)^{6 - 0} + ^{6}{}{C}_1 \left( \frac{1}{15} \right)^1 \left( \frac{14}{15} \right)^{6 - 1} + ^{6}{}{C}_2 \left( \frac{1}{15} \right)^2 \left( \frac{14}{15} \right)^{6 - 2} \right\}

1 - \left\{ \left( \frac{14}{15} \right)^6 + \frac{6}{15} \left( \frac{14}{15} \right)^5 + \frac{1}{15} \left( \frac{14}{15} \right)^4 \right\}

Question 10. If getting 5 or 6 in a throw of an unbiased die is a success and the random variable X denotes the number of successes in six throws of the die, find P (X ≥ 4).

Solution:

Let us consider X be the number of successes. That is getting 5 or 6 in a throw of die in 6 throws.

Now, a binomial distribution X follows with n = 6

Here, p = of getting 5 or 6 = 1/6 + 1/6 = 1/3

 And q = 1 – p = 2/3

P(X = r) = ^{6}{}{C}_r \left( \frac{1}{3} \right)^r \left( \frac{2}{3} \right)^{6 - r}

P(X > 4) = P(X = 4) + P(X = 5) + P(X = 6)

^{6}{}{C}_4 \left( \frac{1}{3} \right)^4 \left( \frac{2}{3} \right)^{6 - 4} +^{6}{}{C}_5 \left( \frac{1}{3} \right)^5 \left( \frac{2}{3} \right)^{6 - 5} + ^{6}{}{C}_6 \left( \frac{1}{3} \right)^6 \left( \frac{2}{3} \right)^{6 - 6}

= (1/36) (60 + 12 + 1)

= 73/729

Question 11. Eight coins are thrown simultaneously. Find the chance of obtaining at least six heads.

Solution:

Let us consider X denotes the number of heads in tossing 8 coins.

Now, a binomial distribution X follows with n = 8.

Here, p = 1/2 and q = 1/2

P(X = r) = ^{8}{}{C}_r \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^{8 - r}

8Cr (1/2)8

So, the probability of getting at least 6 heads is 

P(X > 6) = P(X = 6) + P(X = 7) + P(X = 8)

^{8}{}{C}_6 \left( \frac{1}{2} \right)^8 + ^{8}{}{C}_7 \left( \frac{1}{2} \right)^8 + ^{8}{}{C}_8 \left( \frac{1}{2} \right)^8

= 1/28(28 + 8 + 1)

= 37/256

Question 12. Five cards are drawn successively with replacement from a well-shuffled pack of 52 cards.

(i) What is the probability that all the five cards are spades?

Solution:

Let us consider X be the number of spade cards when 5 cards are drawn with replacement.

Now, a binomial distribution X follows with n = 5.

p = 13/52 = 1/4 and q = 1 – p = 3/4

P(X = r) = ^{5}{}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^{5 - r}

P(All cards are spades) = P(X = 5)

^{5}{}{C}_5 \left( \frac{1}{4} \right)^5 \left( \frac{3}{4} \right)^0

= 1/1024

(ii) What is the probability that only 3 cards are spades? 

Solution:

Let us consider X be the number of spade cards when 5 cards are drawn with replacement.

Now, a binomial distribution X follows with n = 5. 

Here, p = 13/52 = 1/4 and q = 1 – p = 3/4

P(X = r) = ^{5}{}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^{5 - r}

P(only 3 cards are spades) = P(X = 3)

^{5}{}{C}_3 \left( \frac{1}{4} \right)^3 \left( \frac{3}{4} \right)^2

= (1/1024) (90)

= 45/512

(iii) What is the probability that none is a spade?

Solution:

Let us consider X be the number of spade cards when 5 cards are drawn with replacement.

Now, a binomial distribution X follows with n = 5.

Here p = 13/52 = 1/4 and q = 1 – p = 3/4.

P(X = r) = ^{5}{}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^{5 - r}

P(none is a spade) = P(X = 0)

^{5}{}{C}_0 \left( \frac{1}{4} \right)^0 \left( \frac{3}{4} \right)^5

= 243/1024

Question 13. A bag contains 7 red, 5 white, and 8 black balls. Four balls are drawn one by one with replacement.

(i) What is the probability that none is white?

Solution:

Let us consider X denotes the number of white balls drawn when 4 balls are drawn with replacement.  

Now, a binomial distribution X follows with n = 4.

Here, the probability for a white ball(p) = 5/20 = 1/4

And q = 1 – p = 3/4

P(X = r) = ^{4}{}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^{4 - r}

So, the probability that none is white is

P(X = 0) = ^{4}{}{C}_0 \left( \frac{1}{4} \right)^0 \left( \frac{3}{4} \right)^{4 - 0}

= 81/256

(ii) What is the probability that none is white?

Solution:

Let us consider X denotes the number of white balls drawn when 4 balls are drawn with replacement.  

Now, a binomial distribution X follows with n = 4.

Here, the probability for a white ball(p) = 5/20 = 1/4

And q = 1 – p = 3/4

P(X = r) = ^{4}{}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^{4 - r}

so, the probability that none is white is 

P(X = 0) = ^{4}{}{C}_0 \left( \frac{1}{4} \right)^0 \left( \frac{3}{4} \right)^{4 - 0}

= 81/256

(iii) What is the probability that any two are white?

Solution:

Let us consider X denotes the number of white balls drawn when 4 balls are drawn with replacement.  

Now, a binomial distribution X follows with n = 4.

So, the probability for a white ball(p) = 5/20 = 1/4

And q = 1 – p = 3/4

P(X = r) = ^{4}{}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^{4 - r}

So, the probability that any two are white is 

P(X = 2) = ^{4}{}{C}_2 \left( \frac{1}{4} \right)^2 \left( \frac{3}{4} \right)^{4 - 2}

= 54/256

= 27/128

Question 14. A box contains 100 tickets, each bearing one of the numbers from 1 to 100. If 5 tickets are drawn successively with replacement from the box, find the probability that all the tickets bear numbers divisible by 10.

Solution:

Let us consider X denotes the variable representing number on the ticket bearing a number divisible by 10 out of the 5 tickets drawn.

Now, a binomial distribution X follows with n = 5.

Here, the probability of getting a ticket bearing number divisible by 10(p) = 10/100

= 1/10

And q = 1 – p = 9/10

P(X = r) = ^{5}{}{C}_r \left( \frac{1}{10} \right)^r \left( \frac{9}{10} \right)^{5 - r}

So, the probability that all the tickets bear numbers divisible by 10 is

P(X = 5) = ^{5}{}{C}_5 \left( \frac{1}{10} \right)^5 \left( \frac{9}{10} \right)^{5 - 5}

= (1/10)5 (9/10)0

= (1/10)5

Question 15. A bag contains 10 balls, each marked with one of the digits from 0 to 9. If four balls are drawn successively with replacement from the bag, what is the probability that none is marked with the digit 0?

Solution:

Let us consider X denotes the number of balls marked with the digit 0 when 4 balls are drawn successfully with replacement.

Now, a binomial distribution X follows with n = 4.

So, the probability that a ball randomly drawn bears digit 0(p) = 1/10

And q = 1 – p = 9/10

P(X = r) = ^{4}{}{C}_r \left( \frac{1}{10} \right)^r \left( \frac{9}{10} \right)^{4 - r}

P(none bears the digit 0) = P(X = 0)

^{4}{}{C}_0 \left( \frac{1}{10} \right)^0 \left( \frac{9}{10} \right)^{4 - 0}

= (9/10)4

Question 16. In a large bulk of items, 5 percent of the items are defective. What is the probability that a sample of 10 items will include not more than one defective item?

Solution:

Let us consider X be the number of defective items in a sample of 10 items.

Now, a binomial distribution X follows with n = 10.

And the probability of defective items(p) = 5% = 0.05 

And q = 1 – p = 0.95

P(X = r) = 10Cr (0. 05)r (0.95)10-r

So, the probability(sample of 10 items will include not more than one defective item) is 

P(X < 1) = P(X = 0) + P(X = 1) 

= 10C0 (0.05)0 (0.95 ){0-0 10C1 (0.05)1 (0.95)10-1

= (0.95)9 (0.95 + 0.5)

= 1.45 (0.95)9

Question 17. The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. 

(i) Find the probability that out of 5 such bulbs none will fuse after 150 days of use. 

Solution:

Let us consider X be the number of bulbs that fuse after 150 days.

Now, a binomial distribution X follows with n = 5.

Here, p = 0.05 and  q = 0.95

Or p = 1/20 and q = 19/20

P(X = r) = ^{5}{}{C}_r \left( \frac{1}{20} \right)^r \left( \frac{19}{20} \right)^{5 - r}

So, the probability (none will fuse after 150 days of use) is 

P(X = 0) = ^{5}{}{C}_0 \left( \frac{1}{20} \right)^0 \left( \frac{19}{20} \right)^{5 - 0}

= (19/20)5

(ii) Find the probability that out of 5 such bulbs, not more than one will fuse after 150 days of use. 

Solution:

Let us consider X denotes the number of bulbs that fuse after 150 days.

Now, a binomial distribution X follows with n = 5.

Here p = 0.05 and q = 0.95.

Or p = 1/20 and q = 19/20

P(X = r) = ^{5}{}{C}_r \left( \frac{1}{20} \right)^r \left( \frac{19}{20} \right)^{5 - r}

So, the probability (not more than 1 will fuse after 150 days of use) is

P(X < 1) = P(X = 0) + P(X = 1)

\left( \frac{19}{20} \right)^5 + 5 C_1 \left( \frac{1}{20} \right)^1 \left( \frac{19}{20} \right)^{5 - 1}

\left( \frac{19}{20} \right)^4 \left\{ \frac{19}{20} + \frac{5}{20} \right\}

= (6/5) (19/20)4

(iii) Find the probability that out of 5 such bulbs more than one will fuse after 150 days of use 

Solution:

Let us consider X denotes the number of bulbs that fuse after 150 days.

Now, a binomial distribution X follows with n = 5.

Here p = 0.05 and q = 0.95

Or p = 1/20 and q = 19/20

P(X = r) = ^{5}{}{C}_r \left( \frac{1}{20} \right)^r \left( \frac{19}{20} \right)^{5 - r}

So, the probability(more than one will fuse after 150 days of use) is

P(X > 1) = 1 – P(X < 1)

= 1 – (6/5) (19/20)4

(iv) Find the probability that out of 5 such bulbs at least one will fuse after 150 days of use. 

Solution:

Let us consider X denotes the number of bulbs that fuse after 150 days.

Now, a binomial distribution X follows with n = 5.

Here p = 0.05 and q = 0.95

Or p = 1/20 and q = 19/20

P(X = r) = ^{5}{}{C}_r \left( \frac{1}{20} \right)^r \left( \frac{19}{20} \right)^{5 - r}

So, the probability(at least one will fuse after 150 days of use) is

P(X > 1) = 1 – P(X = 0)

= 1 – (19/20)5

Question 18. Suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed?

Solution:

Let us consider X denotes the number of people that are right-handed in the sample of 10 people.

Now, a binomial distribution X follows with n = 10.

Here p = 90 % = 90/100 = 0.9

And q = 1 – p = 0.1

P(X = r) = 10Cr (0.9)r (0.1)10-r

So, the probability that at most 6 are right – handed is

P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)

= 1 – {P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)}

1 - \sum^{10}_{r = 7}{^{10}{}{C}_r} (0 . 9 )^r (0 . 1 )^{10 - r}

Question 19. A bag contains 7 green, 4 white, and 5 red balls. If four balls are drawn one by one with replacement, what is the probability that one is red?

Solution:

Let us consider X be the number of red balls drawn from 16 balls with replacement.

So, a binomial distribution follows by X with n = 4.

Here, p = 5/16 and q = 1 – p = 11/16

P(X = r) = ^{4}{}{C}_r \left( \frac{5}{16} \right)^r \left( \frac{11}{16} \right)^{4 - r}

P(One ball is red) = P(X = 1)

^{4}{}{C}_1 \left( \frac{5}{16} \right)^1 \left( \frac{11}{16} \right)^{4 - 1}

= 4 (5/16) (11/16)3

= (5/4) (11/16)3

Question 20. A bag contains 2 white, 3 red, and 4 blue balls. Two balls are drawn at random from the bag. If X denotes the number of white balls among the two balls drawn, describe the probability distribution of X.

Solution:

Let X denote the number of white balls when 2 balls are drawn from the bag.

So, X follows a distribution with values 0, 1, or 2.

P(X = 0) = P(All balls non – white) 

\frac{^{7}{}{C}_2}{^{9}{}{C}_2}

= 42/72

= 21/36

P(X = 1) = P ( Ist ball white and IInd ball non – white)

\frac{^{7}{}{C}_1 ^{2}{}{C}_1}{^{9}{}{C}_2}

= 14/36

So, P(X = 2) = P(Both balls white) 

\frac{^{2}{}{C}_2}{^{9}{}{C}_2}

= 1/36

So, the tabular form is:

X012
P(X)21/3614/361/36

Question 21. An urn contains four white and three red balls. Find the probability distribution of the number of red balls in three draws with replacement from the urn.

Solution:

Given that three balls are drawn with a replacement, the number of white balls.

So, a binomial distribution follows by X with n = 3.

Here p = 3/7 and q = 4/7

P(X = r) = ^{3}{}{C}_r \left( \frac{3}{7} \right)^r \left( \frac{4}{7} \right)^{3 - r}     , r = 0, 1, 2, 3

P(X = 0) = ^{3}{}{C}_0 \left( \frac{3}{7} \right)^0 \left( \frac{4}{7} \right)^{3 - 0}

= 64/343

P(X = 1) = ^{3}{}{C}_1 \left( \frac{3}{7} \right)^1 \left( \frac{4}{7} \right)^{3 - 1}

= 144/343

P(X = 2) = ^{3}{}{C}_2 \left( \frac{3}{7} \right)^2 \left( \frac{4}{7} \right)^{3 - 2}

= 108/343

P(X = 3) = ^{3}{}{C}_3 \left( \frac{3}{7} \right)^3 \left( \frac{4}{7} \right)^{3 - 3}

= 27/343

So, the tabular form is:

X0123
P(X)64/343144/343108/34327/343

Question 22. Find the probability distribution of the number of doublets in 4 throws of a pair of dice.

Solution:

Let us considered X denotes the number of doublets in 4 throws of a pair of dice.

So, a binomial distribution follows by X with n = 4.

Here p = No of getting (1, 1)(2, 2) . . . (6, 6) 

= 6/36

= 1/6

And q = 1 – p = 5/6

P(X = r) = ^{4}{}{C}_r \left( \frac{1}{6} \right)^r \left( \frac{5}{6} \right)^{4 - r}    , r = 0, 1, 2, 3, 4

P(X = 0) = ^{4}{}{C}_0 \left( \frac{1}{6} \right)^0 \left( \frac{5}{6} \right)^{4 - 0}

= 625/1296

P(X = 1) = ^{4}{}{C}_1 \left( \frac{1}{6} \right)^1 \left( \frac{5}{6} \right)^{4 - 1}

= 500/1296

P(X = 2) = ^{4}{}{C}_2 \left( \frac{1}{6} \right)^2 \left( \frac{5}{6} \right)^{4 - 2}

= 150/1296

P(X = 3) = ^{4}{}{C}_3 \left( \frac{1}{6} \right)^3 \left( \frac{5}{6} \right)^{4 - 3}

= 20/1296

P(X = 4) = ^{4}{}{C}_4 \left( \frac{1}{6} \right)^4 \left( \frac{5}{6} \right)^{4 - 4}

= 1/1296

So, the distribution is:

X01234
P(X)625/1296500/1296150/129620/12961/1296

Question 23. Find the probability distribution of the number of sixes in three tosses of a die.

Solution:

Let us considered X denotes the number of 6 in 3 tosses of a die.

So, a binomial distribution follows by X with n = 3.

Here p = 1/6, q = 1 – p = 5/6

P(X = r) = ^{3}{}{C}_r \left( \frac{1}{6} \right)^r \left( \frac{5}{6} \right)^{3 - r}    , r = 0, 1, 2, 3

So, P(X = 0) = ^{3}{}{C}_0 \left( \frac{1}{6} \right)^0 \left( \frac{5}{6} \right)^{3 - 0}

= 125/216

P(X = 1) = ^{3}{}{C}_1 \left( \frac{1}{6} \right)^1 \left( \frac{5}{6} \right)^{3 - 1}

= 75/216

P(X = 2) = ^{3}{}{C}_2 \left( \frac{1}{6} \right)^2 \left( \frac{5}{6} \right)^{3 - 2}

= 15/216

P(X = 3) = ^{3}{}{C}_3 \left( \frac{1}{6} \right)^3 \left( \frac{5}{6} \right)^{3 - 3}

= 1/216

So, the distribution is:

X0123
P(X)125/21675/21615/2161/216

Question 24. A coin is tossed 5 times. If X is the number of heads observed, find the probability distribution of X.

Solution:

Let us considered X denotes the number of heads in 5 tosses. 

So, a binomial distribution follows by X with n = 5.

Here p = 1/2 and q = 1/2

P(X = r) = ^{5}{}{C}_r \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^{5 - r}    , r = 0, 1, 2, 3, 4, 5

5C(2)5

P(X = 0) = 5C0 (2)5

= 1/32

P(X = 1) = 5C1 (2)5

= 5/32

P(X = 2) = 5C2 (2)5

= 10/32

P(X = 3) = 5C3 (2)5

= 10/32

P(X = 4) = 5C4 (2)5

= 5/32

P(X = 5) = 5C(2)5

= 1/32

So, the distribution is:

X012345
P(X)1/325/3210/3210/325/321/32

Question 25. An unbiased die is thrown twice. A success is getting a number greater than 4. Find the probability distribution of the number of successes.

Solution:

Let us consider X be the getting a number greater than 4 .

So, a binomial distribution follows by X with n = 2.

Here p = P(X > 4) = P(X = 5 or 6)

= 1/6 + 1/6

= 1/3

And q = 1 – p = 2/3

P(X = r) = ^{2}{}{C}_r \left( \frac{1}{3} \right)^r \left( \frac{2}{3} \right)^{2 - r}   , r = 0, 1, 2

So, the distribution is:

X012
P(X)4/94/91/9

Question 26. A man wins a rupee for head and loses a rupee for tail when a coin is tossed. Suppose that he tosses once and quits if he wins but tries once more if he loses on the first toss. Find the probability distribution of the number of rupees the man wins.

Solution:

Let us consider X be the number of rupees the man wins.

So, first we assume that he gets head in the first toss.

So, the probability would be 1/2. Also, he wins Rs.1 rupee.

Now, the second possibility is that he gets a tail in the first toss.

Then he tosses again. Suppose he obtain ahead in the second toss.

Then, he wins Rs 1 rupee in the second toss but loses Rs 1 rupee in the first toss.

So, the money he won = Rs 0

So, the probability for winning Rs.0 is 

= (1/2) (1/2)

= 1/4

Now, the third possibility is obtaining tail in the first toss and also tail in the second toss

Then, the money that he would win = -2 (As he loses Rs 2)

So, the probability for the third possibility = (1/2) (1/2)

= 1/4

So, the distribution is:

X012
P(X)1/21/41/4

Question 27. Five dice are thrown simultaneously. If the occurrence of 3, 4, or 5 in a single die is considered a success, find the probability of at least 3 successes.

Solution:

Let us consider X be the occurrence of 3,4 or 5 in a single die. 

So, a binomial distribution follows by X with n = 5.

Let us assume the probability of getting 3, 4 or 5 in a single die is p

Here p = 3/6 = 1/2

And q = 1 – 1/2 = 1/2

P(X = r) = ^{5}{}{C}_r \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^{5 - r}

P(at least 3 successes) = P(X > 3)

= P(X = 3) + P(X = 4) + P(X = 5)

^{5}{}{C}_3 \left( \frac{1}{2} \right)^3 \left( \frac{1}{2} \right)^{5 - 3} + ^{5}{}{C}_4 \left( \frac{1}{2} \right)^4 \left( \frac{1}{2} \right)^{5 - 4} +^{5}{}{C}_5 \left( \frac{1}{2} \right)^5 \left( \frac{1}{2} \right)^{5 - 5}

\frac{^{5}{}{C}_3 + ^{5}{}{C}_4 + ^{5}{}{C}_5}{2^5}

= 16/32

= 1/2

Question 28. The items produced by a company contain 10% defective items. Show that the probability of getting 2 defective items in a sample of 8 items is \frac{28 \times 9^6}{{10}^8}   .

Solution:

Let us consider X be the number of defective items in the items produced by the company.

So, a binomial distribution follows by X with n = 8.

Here, p = 10 % = 10/100 = 1/10

And q = 1 – p = 9/10

Hence, the distribution is given by,

P(X = r) = ^{8}{}{C}_r \left( \frac{1}{10} \right)^r \left( \frac{9}{10} \right)^{8 - r}

So, the probability of getting 2 defective items is

P(X = 2) = ^{8}{}{C}_2 \left( \frac{1}{10} \right)^2 \left( \frac{9}{10} \right)^{8 - 2}

\frac{28 \times 9^6}{{10}^8}

Hence proved.

Question 29. A card is drawn and replaced in an ordinary pack of 52 cards. How many times must a card be drawn so that 

(i) there is at least an even chance of drawing a heart 

Solution:

Let us consider X be the probability of drawing a heart from a deck of 52 cards. 

So, we get

Here, p = 13/52 = 1/4

And q = 1 – p = 1 – 1/4 = 3/4

Let us assume the card be drawn n times. 

So, 

P(X = r) = ^{n}{}{C}_r p^r q^{n - r}

Let us consider X be the number of hearts drawn from a pack of 52 cards.

So, the smallest value of n for which P(X=0) is less than 1/4

i.e., P(X = 0) < 1/4

^{n}{}{C}_0 \left( \frac{1}{4} \right)^0 \left( \frac{3}{4} \right)^{n - 0} < \frac{1}{4}

=> \left( \frac{3}{4} \right)^n < \frac{1}{4}

Now, put n = 1, (3/4)1 not less than 1/4

n = 2, (3/4)2 not less than 1/4

n = 3, (3/4)3 not less than 1/4

So, smallest value of n = 3.

(ii) the probability of drawing a heart is greater than 3/4

Solution:

Given that the probability of drawing a heart > 3/4.

so, 1 – P(X = 0) > 3/4

1 -^{n}{}{C}_0 \left( \frac{1}{4} \right)^0 \left( \frac{3}{4} \right)^{n - 0} > \frac{3}{4}
1 - \left( \frac{3}{4} \right)^n > \frac{3}{4}
1 - \frac{3}{4} > \left( \frac{3}{4} \right)^n
\frac{1}{4} > \left( \frac{3}{4} \right)^n

For n = 1, (3/4)1 not less than 1/4.

n = 2, (3/4)2 not less than 1/4

n = 3, (3/4)3 not less than 1/4

n = 4, (3/4)4 not less than 1/4

n = 5, (3/4)5 not less than 1/4

So, card must be drawn 5 times.

Question 30. The mathematics department has 8 graduate assistants who are assigned to the same office. Each assistant is just as likely to study at home as in office. How many desks must there be in the office so that each assistant has a desk at least 90% of the time?

Solution:

Let us consider k denotes the number of desks and X denotes the number of graduate assistants in the office.

So, a binomial distribution follows by X with n = 8.

Here p = 1/2 and q = 1/2.

So, 

=> P(X < k) > 90%  

=> P\left( X < k \right) > 0.90

=> P\left( X > k \right) < 0.10

=> P(X = k + 1, k + 2, . . . . 8) < 0 . 10

Therefore, P(X > 6) = P(X = 7 or X = 8)

^{8}{}{C}_7 \left( \frac{1}{2} \right)^8 + ^{8}{}{C}_8 \left( \frac{1}{2} \right)^8 = 0.04

Now, P(X > 5) = P(X = 6, X = 7 or X = 8) = 0.15

P(X > 6) < 0.10

Hence, if there are 6 desks then there is at least 90% chance for every graduate to get a desk.

Question 31. An unbiased coin is tossed 8 times. Find, by using a binomial distribution, the probability of getting at least 6 heads.

Solution:

Let us considered X be the number of heads in tossing the coin 8 times.

So, a binomial distribution follows by X with n = 8.

Here p = 1/2 and q = 1/2

Hence, 

P(X = r) = ^{8}{}{C}_r \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^{8 - r}   , r = 0, 1, 2, 3, 4, 5, 6, 7, 8

So, the required probability is 

P(X > 6) = P(X = 6) + P(X = 7) + P(X = 8)

\frac{^{8}{}{C}_6 + ^{8}{}{C}_7 + ^{8}{}{C}_8}{2^8}

\frac{28 + 8 + 1}{256}

= 37/256

Question 32. Six coins are tossed simultaneously. Find the probability of getting:

(i) 3 heads

Solution:

Let us considered X be the number of heads obtained in tossing 6 coins.

So, a binomial distribution follows by X with n = 6.

Here p = 1/2 and q = 1/2

Hence, 

P(X = r) = ^{6}{}{C}_r \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^{6 - r}   , r = 0, 1, 2, 3, 4, 5, 6

\frac{^{6}{}{C}_r}{2^6}

P(getting  3 heads) = P(X = 3)

\frac{^{6}{}{C}_3}{2^6}

= 20/64

= 5/16

(ii) no heads

Solution:

P(getting no head) = P(X = 0)

\frac{^{6}{}{C}_0}{2^6}

= (1/2)6

= 1/64

(iii) at least one head

Solution:

P(getting at least 1 head) = P(X > 1)

= 1 – P(X = 0)

= 1 – 1/64

= 63/64

Question 33. Suppose that a radio tube inserted into a certain type of set has a probability of 0.2 of functioning more than 500 hours. If we test 4 tubes at random what is the probability that exactly three of these tubes function for more than 500 hours?

Solution:

Let us considered X be the number of tubes that function for more than 500 hours.  

So, a binomial distribution follows by X with n = 4.

Let us considered p be the probability that the tubes function more than 500 hours.

Here , p = 0.2, q = 0.8

Hence, 

P(X = r) = 4Cr (0.2)r (0.8)4-r, r = 0, 1, 2, 3, 4

So, the required probability is 

P(X = 3) = 4 (0.2)3 (0.8)

= 0.0256

Question 34. The probability that a certain kind of component will survive a given shock test is 3/4. 

(i) Find the probability that among 5 components tested exactly 2 will survive.

Solution:

Let us considered X be the number of components that survive shock.

So, a binomial distribution follows by X with n = 5.

Let us considered p be the probability that a certain kind of 

component will survive a given shock test.  

So, p = 3/4 and q = 1/4

Hence, 

P(X = r) = ^{5}{}{C}_r \left( \frac{3}{4} \right)^r \left( \frac{1}{4} \right)^{5 - r}    , r = 0, 1, 2, 3, 4, 5

P(exactly 2 will survive} ) = P(X = 2) 

^{5}{}{C}_2 \left( \frac{3}{4} \right)^2 \left( \frac{1}{4} \right)^{5 - 2}

\frac{10 \times 9}{1024}

= 0.0879

(ii) Find the probability that among 5 components tested at most 3 will survive.

Solution:

Let us considered X be the number of components that survive shock.

So, a binomial distribution follows by X with n = 5.

Let us considered p be the probability that a certain kind of 

component will survive a given shock test.

So, p = 3/4 and q = 1/4

Hence, 

P(X = r) = ^{5}{}{C}_r \left( \frac{3}{4} \right)^r \left( \frac{1}{4} \right)^{5 - r}   , r = 0, 1, 2, 3, 4, 5

P(at most 3 will survive) = P(X < 3)

=  P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

^{5}{}{C}_{0} \left( \frac{3}{4} \right)^0 \left( \frac{1}{4} \right)^{5 - 0} + ^{5}{}{C}_1 \left( \frac{3}{4} \right)^1 \left( \frac{1}{4} \right)^{5 - 1} + ^{5}{}{C}_2 \left( \frac{3}{4} \right)^2 \left( \frac{1}{4} \right)^{5 - 2} + ^{5}{}{C}_3 \left( \frac{3}{4} \right)^3 \left( \frac{1}{4} \right)^{5 - 3}

\left( \frac{1}{4} \right)^5 + 5\left( \frac{3}{4} \right) \left( \frac{1}{4} \right)^4 + 10 \left( \frac{3}{4} \right)^2 \left( \frac{1}{4} \right)^3 + 10 \left( \frac{3}{4} \right)^3 \left( \frac{1}{4} \right)^2

\frac{1 + 15 + 90 + 270}{1024}

= 376/1024

= 0.3672

Question 35. Assume that the probability that a bomb dropped from an aeroplane will strike a certain target is 0.2. If 6 bombs are dropped, find the probability:

(i) that exactly 2 will strike the target.

Solution:

Let us considered X be the number of bombs that hit the target and 

p be the probability that a bomb dropped from an aeroplane will strike the target.

So, a binomial distribution follows by X with n  = 6

p = 0.2 and q = 0.8

Hence, 

P(X = r) = ^{6}{}{C}_r \left( 0 . 2 \right)^r \left( 0 . 8 \right)^{6 - r}

P(exactly 2 will strike the target) = P(X = 2)

^{6}{}{C}_2 (0 . 2 )^2 (0 . 8 )^4

= 0.2458

(ii) that at least 2 will strike the target

Solution:

Let us considered X be the number of bombs that hit the target and

p be the probability that a bomb dropped from an aeroplane will strike the target.

So, a binomial distribution follows by X with n = 6.

p = 0.2 and q = 0.8

Hence, 

P(X = r) = ^{6}{}{C}_r \left( 0 . 2 \right)^r \left( 0 . 8 \right)^{6 - r}

P(at least 2 will strike the target) = P(X > 2)

= 1 – [P(X = 0) + P(X = 1)]

= 1 – (0.8)6 – 6 (0.2) (0.8)5

= 1 – 0.2621 – 0.3932

= 0 . 3447

Question 36. It is known that 60% of mice inoculated with serum are protected from a certain disease. If 5 mice are inoculated, find the probability that: 

(i) none contract the disease.

Solution:

Let us considered X be the number of mice that contract the disease and

p be the probability of mice that contract the disease.

So, a binomial distribution follows by X with n  = 5.

 p = 0.4 and q = 0.6

Hence, 

P(X = r) = ^{5}{}{C}_r \left( 0 . 4 \right)^r \left( 0 . 6 \right)^{5 - r}   , r = 0, 1, 2, 3, 4, 5

P(X = 0) = ^{5}{}{C}_0 \left( 0 . 4 \right)^0 \left( 0 . 6 \right)^{5 - 0}

= (0.6)5

= 0.0778

(ii) more than 3 contracts the disease.

Solution:

Let us considered X be the number of mice that contract the disease and

 p be the probability of mice that contract the disease.

So, a binomial distribution follows by X with n  = 5.

 p = 0.4 and q = 0.6

Hence, 

P(X = r) = ^{5}{}{C}_r \left( 0 . 4 \right)^r \left( 0 . 6 \right)^{5 - r}   , r = 0, 1, 2, 3, 4, 5

P(X > 3) = P(X = 4) + P(X = 5)

^{5}{}{C}_4 \left( 0 . 4 \right)^4 \left( 0 . 6 \right)^{5 - 4} +^{5}{}{C}_5 \left( 0 . 4 \right)^5 \left( 0 . 6 \right)^{5 - 5}

= 0.0768 + 0.01024

= 0.08704

Question 37. An experiment succeeds twice as often as it fails. Find the probability that in the next 6 trials there will be at least 4 successes.
Solution:

Let us consider X be the number of successes out of 6 experiment. Also, p denotes the probability of success and q denotes the probability of failure.
According to the question, successes are twice failures.
So,  p = 2q
Also, p + q = 1
=> 3q = 1
=> q = 1/3
And p = 1 – 1/3 = 2/3
So, the binomial distribution(n) = 6
Hence, the probability of getting r success is,
P(X = r) = ^{6}{}{C}_r \left( \frac{2}{3} \right)^r \left( \frac{1}{3} \right)^{6 - r}  , r = 0, 1, 2 . . . . . 6
P(at least 4 successes) = P(X > 4)
= P(X = 4) + P(X = 5) + P(X = 6)
^{6}{}{C}_4 \left( \frac{2}{3} \right)^4 \left( \frac{1}{3} \right)^{6 - 4} + ^{6}{}{C}_5 \left( \frac{2}{3} \right)^5 \left( \frac{1}{3} \right)^{6 - 5} + ^{6}{}{C}_6 \left( \frac{2}{3} \right)^6 \left( \frac{1}{3} \right)^{6 - 6}
\frac{15( 2^4 ) + 6(32) + 64}{3^6}
\frac{240 + 192 + 64}{729}
= 496/729
Question 38. In a hospital, there are 20 kidney dialysis machines and the chance of any one of them to be out of service during a day is 0.02. Determine the probability that exactly 3 machines will be out of service on the same day.
Solution:
Let us considered X be the number of machines out of service during a day. Also, p be the probability of any machine out of service during a day and q be the probability that machine will be in service on the same day.
Then, X has a binomial distribution with n = 20.
Hence, p = 0.02 and q = 0.98
Therefore, 
P(X = r) = ^{20}{}{C}_r \left( 0 . 02 \right)^r \left( 0 . 98 \right)^{20 - r}  , r = 0, 1, 2 . . . . . 20
The probability of exactly 3 machines will be out of the service on the same day is
P(X = 3) = ^{20}{}{C}_3 \left( 0 . 02 \right)^3 \left( 0 . 98 \right)^{20 - 3}
= 1140 (0.000008) (0.7093)
= 0.006469
Question 39. The probability that a student entering a university will graduate is 0.4. 
(i) Find the probability that out of 3 students of the university none will graduate.
Solution:
Let us consider X denotes the number of students that graduate from among 3 students. Also, p denotes the probability that a student entering a university will graduate.
Here, n = 3, p = 0.4 and q = 0.6
Hence, 
P(X = r) = ^{3}{}{C}_r \left( 0 . 4 \right)^r \left( 0 . 6 \right)^{3 - r}   , r = 0, 1, 2, 3
So, the probability that none of the student will graduate is 
P(X = 0) = q
= (0.6)3
= 0.216
(ii) Find the probability that out of 3 students of the university only one will graduate.
Solution:
Let us consider X denotes the number of students that graduate from among 3 students. Also, p denotes the probability that a student entering a university will graduate.
Here , n = 3, p = 0.4 and q = 0.6.
Hence, 
P(X = r) = ^{3}{}{C}_r \left( 0 . 4 \right)^r \left( 0 . 6 \right)^{3 - r}   , r = 0, 1, 2, 3
So, the probability that only one student will graduate is 
P(X = 1) = 3 (0.4) (0.36) 
= 0.432
(iii) Find the probability that out of 3 students of the university all will graduate.
Solution:
Let us consider X denotes the number of students that graduate from among 3 students. Also, p be the probability that a student entering a university will graduate.
Here, n =3, p = 0.4 and q = 0.6
Hence, 
P(X = r) = ^{3}{}{C}_r \left( 0 . 4 \right)^r \left( 0 . 6 \right)^{3 - r}  , r = 0, 1, 2, 3
So, the probability that all the students will graduate is 
P(X = 3) = p
= (0.4)3
= 0.064
Question 40. Ten eggs are drawn successively, with replacement, from a lot containing 10% defective eggs. Find the probability that there is at least one defective egg.
Solution:
Let us consider X denotes the number of defective eggs drawn from 10 eggs. Also, p denotes the probability that a drawn egg is defective.
Then, X has a binomial distribution with n = 10. 
=> p = 10% = 10/100 = 1/10
And q = 1 – p = 9/10
Hence, 
P(X = r) = ^{10}{}{C}_r \left( \frac{1}{10} \right)^r \left( \frac{9}{10} \right)^{10 - r}   , r = 0, 1, 2 . . . . 10
So, the probability of t least one defective egg is 
= P(X > 1) = 1 – P(X = 0)
1 - ^{10}{}{C}_0 \left( \frac{1}{10} \right)^0 \left( \frac{9}{10} \right)^{10 - 0}
= 1 – (9/10)10
Question 41. In a 20-question true-false examination, suppose a student tosses a fair coin to determine his answer to each question. For every head he answers ‘true’ and for every tail, he answers ‘false’. Find the probability that he answers at least 12 questions correctly.
Solution:
Let us consider X be the number of correct answers. Also, p denotes the probability of a correct answer and p be the probability of a correct answer.
So, 
=> p = 1/2
So q = 1 – 1/2 = 1/2
Then, X has a binomial distribution with n = 20.
Hence, 
P(X = r) = ^{20}{}{C}_r \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^{20 - r}   , r = 0, 1, 2, 3 . . . . . . 20
\frac{^{20}{}{C}_r}{2^{20}}
So, the probability that the student answers at least 12 questions correctly is
P(X > 12) = P(X = 12) + P(X = 13) + . . . + P(X = 20)
\frac{^{20}{}{C}_{12} + ^{20}{}{C}_{13} + . . . +^{20}{}{C}_{20}}{2^{20}}
Question 42. Suppose X has a binomial distribution with n = 6 and p = 1/2. Show that X = 3 is the most likely outcome. 
Solution:
According to the question, x has a binomial distribution with n = 6 and p = 1/2.
So, q = 1 – p = 1 – 1/2 = 1/2
Hence, 
P(X = r) = ^{6}{}{C}_r \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^{6 - r}   , r = 0, 1, 2, 3, 4, 5, 6
6Cr (1/2)6
=> P(X = r) = 6Cr (1/2)6
Now, by substituting r = 0, 1, 2, 3, 4, 5 and 6, we get 
P(X = 0) = 6C0 (1/2)6
= 1/64
P(X = r) = 6C1 (1/2)
= 6/64
P(X = r) = 6C2 (1/2)6
= 15/64
P(X = r) = 6C3 (1/2)6
= 20/64
P(X = r) = 6C4 (1/2)6
= 15/64
P(X = r) = 6C5 (1/2)6
= 6/64
P(X = r) = 6C6 (1/2)6
= 1/64
So, the tabular form is:
X
0
1
2
3
4
5
6
P(X)
1/64
6/64
15/64
20/64
15/64
6/64
1/64
Now, on comparing the probabilities, we get that X = 3 is the most likely outcome as P(X = 3) has the greatest value.
Hence proved.
Question 43. In a multiple-choice examination with three possible answers for each of the five questions out of which only one is correct, what is the probability that a candidate would get four or more correct answers just by guessing?
Solution:
Let us consider X denotes the number of right answers in the 5 questions. Also, p denotes the probability of guessing right answer and q denotes the probability of guessing wrong answer.
Here X can take values 0, 1, 2, 3, 4 and 5. So, X has a binomial distribution with n = 5.
Here p = 1/3
And q = 2/3
Hence, 
P(X = r) = ^{5}{}{C}_r \left( \frac{1}{3} \right)^r \left( \frac{2}{3} \right)^{5 - r}   , r = 0, 1, 2, . . . 5
So, the probability that the student guesses 4 or more correct answers
P(X > 4) = P(X = 4) + P(X = 5)
^{5}{}{C}_4 \left( \frac{1}{3} \right)^4 \left( \frac{2}{3} \right)^1 + ^{5}{}{C}_5 \left( \frac{1}{3} \right)^5 \left( \frac{2}{3} \right)^0
\frac{10 + 1}{3^5}
= 11/243
Question 44. A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is 1/100.
(i) What is the probability that he will win a prize at least once.
Solution:
Let us consider X be the number of times the person wins the lottery. Also, p denotes the probability of winning a prize.
Then, X has a binomial distribution with n = 50.
=> p = 1/100
And q = 1 – 1/100 = 99/100
Hence, 
P(X = r) = ^{50}{}{C}_r \left( \frac{1}{100} \right)^r \left( \frac{99}{100} \right)^{50 - r}   , r = 0, 1, 2 . . . 50
So, the probability of winning at least once is 
P(X > 0) = 1 – P(X – 0)
= 1 – (99/100)50
(ii) What is the probability that he will win a prize exactly once.
Solution:
Let us consider X be the number of times the person wins the lottery. Also, p denotes the probability of winning a prize.
Then, X has a binomial distribution with n = 50.
=> p = 1/100
And q = 1 – 1/100 = 99/100
Hence, 
P(X = r) = ^{50}{}{C}_r \left( \frac{1}{100} \right)^r \left( \frac{99}{100} \right)^{50 - r}  , r = 0, 1, 2 . . . 50
So the probability of winning at least twice is 
P(X > 2) = 1 – P(X = 0) – P(X = 1)
1 - \left( \frac{99}{100} \right)^{50} - ^{50}{}{C}_1 \times \frac{1}{100} \times \left( \frac{99}{100} \right)^{49}
1 - \frac{{99}^{49} \times 149}{{100}^{50}}
(iii) What is the probability that he will win a prize at least twice.
Solution:
Let us consider X be the number of times the person wins the lottery. Also, p be the probability of winning a prize.
Then, X has a binomial distribution with n = 50.
=> p = 1/100
And q = 1 – 1/100 = 99/100
Hence, 
P(X = r) = ^{50}{}{C}_r \left( \frac{1}{100} \right)^r \left( \frac{99}{100} \right)^{50 - r}  , r = 0, 1, 2 . . . 50
So, the probability of winning at least twice is 
P(X > 2) = 1 – P(X = 0) – P(X = 1)
1 - \left( \frac{99}{100} \right)^{50} - ^{50}{}{C}_1 \times \frac{1}{100} \times \left( \frac{99}{100} \right)^{49}
1 - \frac{{99}^{49} \times 149}{{100}^{50}}
Question 45. The probability of a shooter hitting a target is 3/4. How many minimum number of times must he/she fire so that the probability of hitting the target at least once is more than 0.99?
Solution:
Let us consider the shooter fire n times and X be the number of times the shooter hits the target.
Then, X has a binomial distribution with p = 3/4 and q = 1/4 such that,
Hence, 
P(X = r) = ^{n}{}{C}_r \left( \frac{3}{4} \right)^r \left( \frac{1}{4} \right)^{n - r}
P(X = r) = ^{n}{}{C}_r \frac{3^r}{4^n}
Given that P (X > 1) > 0.99
=> 1 – P(X = 0) > 0.99
=> 1 – 1/4n > 0.99
=> 1/4n < 0.01
=> 4n > 1/0.01
=> 4n > 100
So, the shooter must fire at least 4 times.
Question 46. How many times must a man toss a fair coin so that the probability of having at least one head is more than 90%?
Solution:
Let us considered the man tosses a fair coin n times and X be the number of heads in n tosses.
As p = 1/2 and q = 1/2, 
So, 
P(X = r) = ^{n}{}{C}_r \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^{n - r}   , r = 0, 1, 2, 3 . . . . n
It is given that P(X > 1) > 0.9
=> 1 – P(X = 0) > 0.9
=> 1 - ^{n}{}{C}_0 \left( \frac{1}{2} \right)^n > 0 . 9″><br>=>(1/2<sup>n</sup>) < 1/10<br>=> 2^n > 10<br>=> n = 4, 5, 6 . . . . <br>So, the man must toss the coin at least 4 times.<br>Question 47. How many times must a man toss a fair coin so that the probability of having at least one head is more than 80%?<br><strong>Solution:</strong><br>Let us consider X denotes the number of heads and n be the minimum number of times that a man must toss a fair coin. <br>So that probability of X ≥ 1 is more than 80 %.<br>Here X has a binomial distribution with p = 1/2 and q = 1/2.<br>P(X = r) = <sup>n</sup>C<sub>r</sub> (1/2)<sup>n</sup><br>We have P(X > 1) = 1 – P(X = 0) <br>= <img loading=
= 1 – (1/2)n
And for P(X > 1) > 80%
=> 1 – 1/2n > 0.80  
=> (1/2n) < 1 – 0.80 = 0.20
=> 2n > 1/0.2 = 5
We know, 22 < 5 while 23 > 5.
So we get n = 3.
So, the man must toss the coin at least 3 times.
Question 48. A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability distribution of the number of successes.
Solution:
Let us consider p be the probability of getting a doublet in a single throw of a pair of dice. Also, X be the number of getting doublets in 4 throws of a pair of dice. 
So, p = 6/36 = 1/6
And q = 1 – p = 1 – 1/6 = 5/6
Then, X has a binomial distribution with n = 4.
Hence, the probability of getting r doublets
P(X = r) = ^{4}{}{C}_r (\frac{1}{6} )^r (\frac{5}{6} )^{4 - r}  , r = 0, 1, 2, 3, 4
If X = 0, then P(X = 0) = ^{4}{}{C}_0 (\frac{1}{6} )^0 (\frac{5}{6} )^{4 - 0}
=> P = (5/6)4
When X = 1, then P = ^{4}{}{C}_1 \left( \frac{1}{6} \right)^1 \left( \frac{5}{6} \right)^{4 - 1}
\frac{2}{3} \left( \frac{5}{6} \right)^3
When X = 2, then P = ^{4}{}{C}_2 \left( \frac{1}{6} \right)^2 \left( \frac{5}{6} \right)^{4 - 2}
\frac{1}{6} \left( \frac{5}{6} \right)^2
When X = 3, then P = ^{4}{}{C}_3 \left( \frac{1}{6} \right)^3 \left( \frac{5}{6} \right)^{4 - 3}
\frac{10}{3} \left( \frac{1}{6} \right)^3
When X = 4, then P = ^{4}{}{C}_4 (\frac{1}{6} )^4 (\frac{5}{6} )^{4 - 4}
= (1/6)4
= 1/1296
Question 49. From a lot of 30 bulbs that include 6 defective bulbs, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.
Solution:
Let us considered X be the number of defective bulbs in a sample of 4 bulbs drawn successively with replacement.
Then, X has a binomial distribution with n = 4.
Here, p = 6/30 = 1/5
And q = 1 – 1/5 = 4/5
Then, 
P(X = r) = ^{4}{}{C}_r \left( \frac{1}{5} \right)^r \left( \frac{4}{5} \right)^{4 - r}   , r = 0, 1, 2, 3, 4
P(X = 0) = (4/5)4
= 256/625
P(X = 1) = 4\left( \frac{1}{5} \right)^1 \left( \frac{4}{5} \right)^3
= 256/625
P(X = 2) = 6 \left( \frac{1}{5} \right)^2 \left( \frac{4}{5} \right)^2
= 96/625
P(X = 3) = 4 \left( \frac{1}{5} \right)^3 \left( \frac{4}{5} \right)^1
= 16/625
P(X = 4) = (1/5)4
= 1/625
X
0
1
2
3
4
P(X)
256/625
256/625
96/625
16/625
1/625
Question 50. Find the probability that in 10 throws of a fair die, a score which is a multiple of 3 will be obtained in at least 8 of the throws. 
Solution:
Let us consider p be the probability of getting a success and q be the probability of getting a failure. Also, X be the number of success in a sample of 10 trials.
p = 2/6 = 1/3
So, q = 1 – p 
= 1 – 1/3
= 2/3
Then X has a binomial distribution with n = 10, p = 1/3 and q = 2/3.
P(X = r) = ^{10}{}{C}_r p^r q^{10 - r}
^{10}{C}_r \left( \frac{1}{3} \right)^r \left( \frac{2}{3} \right)^{\left( 10 - r \right)}  , r = 0, 1, 2, . . . , 10
So, the required probability is 
P(X > 8) = P(X = 8) + P(X = 9) + P(X = 10)
\frac{^{10}{}{C}_8 2^{\left( 10 - 8 \right)}}{3^{10}} + \frac{^{10}{}{C}_9 2^{\left( 10 - 9 \right)}}{3^{10}} + \frac{^{10}{C}_{10} 2^{\left( 10 - 10 \right)}}{3^{10}}
\frac{45 \times 2^2}{3^{10}} + \frac{10 \times 2}{3^{10}} + \frac{1}{3^{10}}
\frac{180 + 20 + 1}{3^{10}}
= 201/310
Question 51. A die is thrown 5 times. Find the probability that an odd number will come up exactly three times. 
Solution:
Let us consider p be the probability of getting an odd number in a trial. Also, X be the number of success in a sample of 5 trials.
So, p = 3/6 = 1/2
Also, q = 1 – p = 1 – 1/2 = 1/2
Then, X has binomial distribution with n = 5 and p = q = 1/2.
P(X = r) = ^{5}{}{C}_r p^r q^{\left( 5 - r \right)}
^{5}{}{C}_r \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^{\left( 5 - r \right) }
5C3 (1/2)5 , where r = 0, 1, 2, 3, 4, 5
So, the probability that an odd number will come up exactly three times. 
P(X = 3) = 5C3 (1/2)5
= 10/32
= 5/16
Question 52. The probability of a man hitting a target is 0.25. He shoots 7 times. What is the probability of his hitting at least twice?
Solution:
Let us consider p be the probability of hitting the target. Also, X be the number of success in a sample of 7 trial.
So, p = = 0.25 = 1/4
Also q = 1 – p = 1 – 1/4 = 3/4
Then, X has a binomial distribution with parameters n = 7 and p = 1/4.
P(X = r) = ^{7}{}{C}_r p^r q^{\left( 7 - r \right)}
^{7}{}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^{\left( 7 - r \right)}
\frac{^{7}{}{C}_r 3^{\left( 7 - r \right)}}{4^7}  , where r  = 0, 1, 2, 3, 4, 5
So, the probability of his hitting at least twice is
P(X > 2) = 1 – [P(X = 0) + P(X = 1)]
1 - \left[ \frac{^{7}{}{C}_0 3^7}{4^7} + \frac{^{7}{}{C}_1 3^6}{4^7} \right]
= 1 – [2187/16384 + 5103/16384]
= 1 – 7290/16384
= 9094/16384
= 4547/8192
Question 53. A factory produces bulbs. The probability that one bulb is defective is 1/50 and they are packed in boxes of 10. From a single box, find the probability: 
(i) that none of the bulbs is defective.
Solution:
Let us consider p be the probability of getting a defective bulb. Also, X be the number of success in a sample of 10 trials.
Here p = 1/50
Also, q = 1 – p = 1 – 1/50 = 49/50
Then, X has a binomial distribution with n = 10 and p = 1/50
P(X = r) = ^{10}{}{C}_r p^r q^{\left( 10 - r \right)}
^{10}{}{C}_r \left( \frac{1}{50} \right)^r \left( \frac{49}{50} \right)^{\left( 10 - r \right)}
\frac{^{10}{}{C}_r {49}^{\left( 10 - r \right)}}{{50}^{10}}  , where r = 0, 1, 2, 3, . . . , 10
So, the probability that none of the bulb is defective
= P(X = 0) = \frac{^{10}{}{C}_0 {49}^{10}}{{50}^{10}}
= 4910/5010
(ii) that exactly two bulbs are defective.
Solution:
Let us consider p be the probability of getting a defective bulb. Also, X be the number of success in a sample of 10 trials.
Here p  = 1/50
Also, q = 1 – p = 1 – 1/50 = 49/50
Then, X has a binomial distribution with n = 10 and p = 1/50
P(X = r) = ^{10}{}{C}_r p^r q^{\left( 10 - r \right)}
^{10}{}{C}_r \left( \frac{1}{50} \right)^r \left( \frac{49}{50} \right)^{\left( 10 - r \right)}
\frac{^{10}{}{C}_r {49}^{\left( 10 - r \right)}}{{50}^{10}}  , where r = 0, 1, 2, 3, . . . , 10
So, the probability that exactly two bulbs are defective
P(X = 2) = \frac{^{10}{}{C}_2 {49}^8}{{50}^{10}}
\frac{45 \times {49}^8}{{50}^{10}}
(iii) that more than 8 bulbs work properly.       
Solution:
Let us consider p be the probability of getting a defective bulb. Also, X be the number of success in a sample of 10 trials.
Here p = 1/50
Also, q = 1 – p = 1 – 1/50 = 49/50
Then, X has a binomial distribution with n = 10 and p = 1/50
P(X = r) = ^{10}{}{C}_r p^r q^{\left( 10 - r \right)}
^{10}{}{C}_r \left( \frac{1}{50} \right)^r \left( \frac{49}{50} \right)^{\left( 10 - r \right)}
\frac{^{10}{}{C}_r {49}^{\left( 10 - r \right)}}{{50}^{10}}  , where r = 0, 1, 2, 3, . . . , 10
So, the probability that more than 8 bulbs work properly is
= P(X < 0)
= P(X = 0) + P(X = 1)
\frac{^{10}{}{C}_0 {49}^{10}}{{50}^{10}} + \frac{^{10}{}{C}_1 {49}^9}{{50}^{10}}
\frac{{49}^{10}}{{50}^{10}} + \frac{10 \times {49}^9}{{50}^{10}}
\frac{{49}^9}{{50}^{10}}\left( 49 + 10 \right)
\frac{59\left( {49}^9 \right)}{\left( {50}^{10} \right)}
Question 54. A box has 20 pens of which 2 are defective. Calculate the probability that out of 5 pens drawn one by one with replacement, at most 2 are defective.
Solution:
Let us consider p be the probability of drawing a defective pen. Also, X be the number of defective pens drawn. 
Then,
=> p = 2/20 = 1/10
And q = 1 – p = 1 – 1/10 = 9/10
Then, X has a binomial distribution with n = 5
So, the probability of drawing r defective pens is
Now, P(X = r) = ^{5}{}{C}_r \left( \frac{1}{10} \right)^r \left( \frac{9}{10} \right)^{5 - r}  , r = 0, 1, 2, 3, 4, 5
So, the probability of drawing at most 2 defective pens
= P(X  ≤ 2)
= P(X = 0) + P(X = 1) + P(X = 2)
^{5}{}{C}_0 \left( \frac{1}{10} \right)^0 \left( \frac{9}{10} \right)^5 +^{5}{}{C}_1 \left( \frac{1}{10} \right)^1 \left( \frac{9}{10} \right)^4 + ^{5}{}{C}_2 \left( \frac{1}{10} \right)^2 \left( \frac{9}{10} \right)^3
\left( \frac{9}{10} \right)^3 \left( \frac{81}{100} + 5 \times \frac{9}{100} + \frac{10}{100} \right)
\frac{729}{1000} \times \frac{136}{100}
= 0.99144

Class 12 RD Sharma Solutions- Chapter 33 Binomial Distribution – Exercise 33.2 | Set 1

  • Last Updated : 03 Mar, 2021

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Question 1. Can the mean of a binomial distribution be less than its variance?

Solution:

Let np be the mean and npq be the variance of a binomial distribution.

So, 

Mean – Variance = np – npq

Mean – Variance = np (1 – q)

Mean – Variance = np.p

Mean – Variance = np2

Since n can never be a negative number and 

p2 will always be a positive number, thus np> 0 

Then, 

Mean – Variance > 0

Mean > Variance

Hence, a mean of a binomial distribution can never be less than its variance.

Question 2. Determine the binomial distribution whose mean is 9 and variance 9/4.

Solution:

We are given mean(np) = 9 and variance(npq) = 9/4.

Solving for the value of q, we will get 

q = \frac{\frac{9}{4}}{9} = \frac{1}{4}

We know, the relation p + q = 1

So, p = 1 – (1/4) = 3/4          -(1) 

Since, np = 9

So, put the value of p from equation(1), we get

n.(3/4) = 9

n = 12

Now, a binomial distribution is given by the relation: nCpr(q)n-r

P(x = r) = 12Cr(3/4)r(1/4)12-r for r = 0,1,2,3,4,….,12

Question 3. If the mean and variance of a binomial distribution are respectively 9 and 6, find the distribution.

Solution:

We are given mean, np = 9 and variance npq = 6.

Solving for the value of q, we will get 

q = 6/9 = 2/3 

We know, the relation p + q = 1

p = 1 – (2/3) = 1/3          -(1)  

Since, np = 9

So, put the value of p from equation(1), we get

n.(1/3) = 9  

n = 27

Now, a binomial distribution is given by the relation: nCpr(q)n-r

P(x = r) = 27Cr(1/3)r(2/3)27-r for r = 0,1,2,3,4,…,27

Question 4. Find the binomial distribution when the sum of its mean and variance for 5 trials is 4.8.

Solution:

Given n = 5 and np + npq = 4.8

np (1 + q) = 4.8

5p (1 + 1 – p) = 4.8

10p -5p2 = 4.8

50p2 – 100p + 48 = 0

Solving for the value of p we will get 

p = 6/5 or p = 4/5

Since, the value of p cannot exceed 1, we will consider p = 4/5.

Therefore, q = 1 – 4/5 = 1/5

Now, a binomial distribution is given by the relation: nCpr(q)n-r

P(x = r) = 5Cr(4/5)r(1/5)5-r for r = 0,1,2,….,5

Question 5. Determine the binomial distribution whose mean is 20 and variance 16.

Solution:

We are given mean, np = 20 and variance npq = 16.

Solving for the value of q, we will get 

q = 16/20 = 4/5 

We know, the relation p + q = 1

p = 1 – 4/5 = 1/5          -(1)  

Since, np = 20

So, put the value of p from equation(1), we get

n.(1/5) = 20

n = 100

Now, a binomial distribution is given by the relation: nCpr(q)n-r

P(x = r) = 100Cr(1/5)r(4/5)100-r for r = 0,1,2,3,4,…,100

Question 6. In a binomial distribution, the sum and product of the mean and the variance are 25/3 and 50/3  respectively. Find the distribution.

Solution:

We are given sum, np + npq = 25/3 

np(1 + q) = 25/3          -(1)  

Product, np x npq = 50/3          -(2)  

Dividing equation(2) by equation(1), we get 

 \frac{np(npq)}{np(1+q)} = (50/3) × 3/25

npq = 2 (1 + q)

np(1 – p) = 2(2 – p)

np = \frac{2(2-p)}{1-p}               -(3)  

On substituting the value of equation(3) in the relation np + npq = 25/3, we get

\frac{2(2-p)}{1-p} + \frac{2(2-p)}{1-p}.q  = 25/3

\frac{2(2-p)}{1-p} + \frac{2(2-p)}{1-p} . (1 – p) = 25/3

\frac{2(2-p)}{1-p} + \frac{2(2-p)}{1-p} (1 + 1 – p) = 25/3

\frac{2(2-p)}{1-p} (2 – p) = 25/3

6p2+ p – 1 = 0

On solving for the value of p, we will get p = 1/3, therefore q = 2/3 

Now, putting value of p and q in the relation np + npq = 25/3

n.(1/3)(1 + (2/3)) = 25/3

n = 15

Now, a binomial distribution is given by the relation: nCpr(q)n-r

P(x = r) = 15Cr(1/3)r(2/3)15-r for r = 0,1,2,3,4,…,15

Question 7. The mean of a binomial distribution is 20 and the standard deviation 4. Calculate the parameters of the binomial distribution.

Solution:

We are given mean, np = 20          -(1) 

Standard deviation, √npq = 4

npq = 16           -(2) 

On dividing the equation (ii) by equation (i), we get

q = 4/5

Therefore, p = 1 – q = 1 – 4/5 = 1/5

Now, since np = 20

n = 20 x 5 = 100

Now, a binomial distribution is given by the relation: nCpr(q)n-r

P(x = r) = 100Cr(1/5)r(4/5)100-r for r = 0,1,2,3,4,…,100

Question 8. If the probability of a defective bolt is 0.1, find the (i) mean and (ii) standard deviation for the distribution of bolts in a total of 400 bolts.

Solution:

We are given n = 400 and q = 0.1, therefore p = 0.9

(i) Mean = np = 400 × 0.9 = 360

(ii) Standard Deviation = √npq =√(400 × 0.9 ×0.1) = 6

Question 9. Find the binomial distribution whose mean is 5 and variance 10/3.

Solution:

We are given mean, np = 5 and variance npq = 10/3.

Solving for the value of q, we will get 

q = \frac{\frac{10}{3}}{5}  = 2/3 

We know, the relation p + q = 1

p = 1 – (2/3) = 1/3 

Since, np = 5

n.(1/3) = 5

n = 15

Now, a binomial distribution is given by the relation: nCpr(q)n-r

P(x = r) = 15Cr(1/3)r(2/3)15-r for r = 0,1,2,3,4,…,15

Question 10. If on an average 9 ships out of 10 arrive safely at ports, find the mean and S.D. of the ships returning safely out of a total of 500 ships.

Solution:

We are given n = 500,

p = 9/10 and thus q = 1/10

Therefore, mean = np = 500 × 0.9 = 450

Standard deviation = √npq = √(450 × 0.1) = 6.71 

Question 11. The mean and variance of a binomial variate with parameters n and p are 16 and 8, respectively. Find P(X = 0), P(X = 1), and P(X ≥ 2).

Solution:

We are given, mean (np) = 16 and variance (npq) = 8

q = 8/16 = 1/2

Therefore, p = 1 – 1/2 = 1/2

Putting the value of p in the relation, np = 16

n = 16 x 2 = 32

Now, a binomial distribution is given by the relation: nCpr(q)n-r

P (x = r) = 32Cr(1/2)r(1/2)32-r for r = 0,1,2,3,4,…,32

Now, P(X = 0) = 32C0(1/2)32 = (1/2)32

Similarly, P(X = 1) = 32C1(1/2)1(1/2)31 = 32 × (1/2)31

Also, P(X ≥ 2) = 1 – P(X = 0) – P(X = 1)

1 – (1/2)32 – 32 × (1/2)32

1 – \frac{33}{2^{32}}

Question 12. In eight throws of a die, 5 or 6 is considered a success. Find the mean number of successes and the standard deviation.

Solution:

We are given n = 8 and p = 2/6 = 1/3 therefore q = 2/3

Now, mean = np = 8 ×(1/3) = 8/3 and

Standard deviation √npq = \sqrt{8×\frac{1}{3}×\frac{2}{3}} = 4/3 

Question 13. Find the expected number of boys in a family with 8 children, assuming the sex distribution to be equally probable.

Solution:

We are given n = 8 and the probability of having a boy or girl is equal, so p = 1/2 and q = 1/2

Therefore, the expected number of boys in a family = np = 8 × 0.5  = 4

Question 14. The probability that an item produced by a factory is defective is 0.02. A shipment of 10,000 items is sent to its warehouse. Find the expected number of defective items and the standard deviation.

Solution:

We are given n = 10,000, also p = 0.02 therefore, q = 1 – 0.02 = 0.98

Now, the expected number of defective items = np = 10000 × 0.02 = 200

The standard deviation = √npq = \sqrt{10000×0.02×0.98}  = √196 = 14

Question 15.  A dice is thrown thrice. A success is 1 or 6 in a throw. Find the mean and variance of the number of successes.
Solution:

Let p denote the success and q denote failure of an event.
Now, the sample space when a dice is thrown is given by S = {1, 2, 3, 4, 5, 6}
Hence, p = 2/6 = 1/3 and q = 1 – 1/3 = 2/3
Therefore, Mean = np = 3 × 1/3 = 1 and Variance = npq = 1 × 2/3 = 2/3
Question 16. If a random variable X follows a binomial distribution with mean 3 and variance 3/2, find P (X ≤ 5).
Solution:
We are given mean (np) = 3 and variance (npq) = 3/2.
Solving for the value of q, \frac{npq}{np} = \frac{\frac{3}{2}}{3}
q = 1/2, hence we can conclude p = 1 – 1/2 = 1/2
Now putting the value of p in relation, np = 3, we get n = 6
We know that a binomial distribution follows the relation:
P(X = r) = nCpr(q)n-r
Therefore, in this case P(X = r) = 6C(1/2)r(1/2)6-r 
P(X = r) = 6C(1/2)6  
We are required to calculate the value for P(X ≤ 5) = 1 – P(X = 6)
P(X ≤ 5) = 1 – 6C(1/2)6 
P(X ≤ 5) = 1 – (1/64) 
P(X ≤ 5) = 63/64 
Question 17. If X follows a binomial distribution with mean 4 and variance 2, find P(X ≥ 5).
Solution:
We are given mean (np) = 4 and variance (npq) = 2.
Solving for the value of q, npq/np = 2/4 
q = 1/2, hence we can conclude p = 1 – 1/2 = 1/2
Now putting the value of p in relation, np = 4, we get n = 8
We know that a binomial distribution follows the relation: P(X = r) = nCpr(q)n-r
Therefore, in this case P(X = r) = 8C(1/2)r(1/2)8-r  
P(X = r) = 8C(1/2)8  
We are required to calculate the value 
P(X ≥ 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)
P(X ≥ 5) = 8C(1/2)8C(1/2)8C(1/2)8C(1/2)8  
P(X ≥ 5) = (1/2)8[8C5 + 8C6 + 8C7 + 8C8]
P(X ≥ 5) = (56 + 28 + 8 + 1)/256
P(X ≥ 5) = 93/256 
Question 18. The mean and variance of a binomial distribution are 4/3 and 8/9 respectively. Find P(X ≥ 1).
Solution:
We are given mean (np) = 4 and variance (npq) = 2
Solving for the value of q, npq/np = \frac{\frac{8}{9}}{\frac{4}{3}}
q = 2/3, hence we can conclude p = 1 – 2/3 = 1/3
Now putting the value of p in relation, np = 4/3, we get n = 4
We know that a binomial distribution follows the relation: P(X = r) = nCpr(q)n-r
Therefore, in this case P(X = r) = 4C(1/3)r(2/3)4-r  
We are required to calculate the value for P(X ≥ 1) = 1 – P(X = 0)
P(X ≥ 1) = 1 – 4C(1/3)0(2/3)4  
P(X ≥ 1) = 1 – 16/81
P(X ≥ 1) = 65/81
Question 19. If the sum of the mean and variance of a binomial distribution for 6 trials is 10/3, find the distribution.
Solution:
Given n = 6 and np + npq = 10/3
np (1 + q) = 10/3
6p (1 + 1 – p) = 10/3
12p – 6p2 = 10/3
18p2 – 36p + 10 = 0
Solving for the value of p we will get p = 1/3 or p = 5/3. 
Since, the value of p cannot exceed 1, we will consider p = 1/3.
Therefore, q = 1 – 1/3 = 2/3
Now, a binomial distribution is given by the relation: nCpr(q)n-r
P(x = r) = 6C(1/3)r(2/3)6-r for r = 0,1,2,….,6
Question 20. A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability distribution of the number of successes and, hence, find its mean.
Solution:
We are given n = 4 and 
a doublet in the throw of a dice occurs when we get (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)
Therefore, the probability of success, p = 6/36 = 1/6, so q = 1 – 1/6 = 5/6
Now, a binomial distribution is given by the relation: nCpr(q)n-r
P(x = r) = 4C(1/6)r(5/6)4-r for r = 0, 1, 2, 3, 4
Hence, the probability distribution is given as:
X
0
1
2
3
4
P(X)
625/1296
500/1296
150/1296
20/1296
1/1296
Mean = 0 × (625/1296) + 1 × (500/1296) + 2 × (150/1296) + 3 × (20/1296) + 0 × (1/1296)
= 864/ 1296
= 2/3
Question 21. Find the probability distribution of the number of doublets in three throws of a pair of dice and find its mean.
Solution:
We are given n = 3 and 
a doublet in the throw of a dice occurs when we get (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)
Therefore, the probability of success, p = 6/36 = 1/6, so q = 1 – 1/6 = 5/6
Now, a binomial distribution is given by the relation: nCpr(q)n-r
P(x = r) = 3C(1/6)r(5/6)3-r for r = 0, 1, 2, 3
Hence, the probability distribution is given as:
X
0
1
2
3
P(X)
125/216
75/216
15/216
1/216
Mean = 0 × (125/216) + 1 × (75/216) + 2 × (15/216) + 3 × (1/216) 
= 108/216
= 1/2
Question 22. From a lot of 15 bulbs which include 5 defective, a sample of 4 bulbs is drawn one by one with replacement. Find the probability distribution of number of defective bulbs. Hence, find the mean of the distribution.
Solution:
Total number of bulbs = 15 and total defective bulbs = 5 
Thus, the probability of getting one defective bulb with replacement, p = 5/15 = 1/3
 Hence, q = 1 – 1/3 = 2/3.
Now, a binomial distribution is given by the relation: nCpr(q)n-r
P(x = r) = 4C(1/3)r(2/3)4-r for r = 0, 1, 2, 3, 4
Hence, the probability distribution is given as:
X
0
1
2
3
4
P(X)
16/81
32/81
24/81
8/81
1/81
Mean = 0 × (16/81) + 1 × (32/81) + 2 × (24/81) + 3 × (8/81) + 4 × (1/81) 
= 108/81
= 4/3
Question 23. A die is thrown three times. Let X be ‘the number of twos seen’. Find the expectation of X.
Solution:
We are given the number of throws, n = 3
Let p denote the probability of getting a 2 in the throw of a dice, then p = 1/6
Therefore, we can conclude 1 = 1 – 1/6 = 5/6
Now, the expectation of X denotes mean therefore, E(X) = np = 3 × 1/6 = 1/2
Question 24. A die is tossed twice. A ‘success’ is getting an even number on a toss. Find the variance of the number of successes.
Solution:
We are given the number of times the coin is tossed, n = 2
Let p denote the probability of getting even number on dice upon throwing which is a success.
Thus, p = 3/6 = 1/2, therefore we can conclude q = 1 – p = 1 – 1/2 = 1/2
Now, the variance is given by npq.
Variance = 2 × 1/2 × 1/2 = 1/2
Question 25. Three cards are drawn successively with replacement from a well-shuffled pack of 52 cards. Find the probability distribution of the number of spades. Hence, find the mean of the distribution. 
Solution:
Number of cards drawn with replacement, n = 3
p = Probability of getting a spade card upon withdrawal = 13/52 = 1/4
Thus, we can conclude, q = 1 – 1/4 = 3/4
Now, a binomial distribution is given by the relation: nCpr(q)n-r
P(x = r) = 3C(1/4)r(3/4)3-r for r = 0, 1, 2, 3 
Hence, the probability distribution is given as:
X
0
1
2
3
P(X)
27/64
27/64
9/64
1/64
Mean = 0 × (27/64) + 1 × (27/64) + 2 × (9/64) + 3 × (1/64) 
= (27 + 18 + 3)/64
= 48/64 
= 3/4
Question 26. An urn contains 3 white and 6 red balls. Four balls are drawn one by one with replacement from the urn. Find the probability distribution of the number of red balls drawn. Also, find mean and variance of the distribution.
Solution:
Let p denote the probability of drawing a red ball which is considered a success, p = 6/9 = 2/3
And the probability of drawing a white ball which is considered a failure, q = 3/9 = 1/3
We have to draw four balls, so n = 4.
Hence, the mean of the probability distribution = np = 4 × 2/3 = 8/3
And variance = npq = 8/3 × 1/3 = 8/9
Now, a binomial distribution is given by the relation: nCpr(q)n-r
P(x = r) = 4C(2/3)r(1/3)4-r for r = 0, 1, 2, 3, 4
Hence, the probability distribution is given as:
X
0
1
2
3
4
P(X)
1/81
8/81
24/81
32/81
16/81
Question 27. Five bad oranges are accidentally mixed with 20 good ones. If four oranges are drawn one by one successively with replacement, then find the probability distribution of number of bad oranges drawn. Hence, find the mean and variance of the distribution.
Solution:
Let p denote the probability of drawing a bad orange which is considered a success, p = 5/25 = 1/5
And the probability of drawing a good orange which is considered a failure, q = 20/25 = 4/5
We have to draw four oranges, so n = 4
Hence, the mean of the probability distribution = np = 4 × 1/5 = 4/5
And the variance = npq = 4/5 × 4/5 = 16/25
Now, a binomial distribution is given by the relation: nCpr(q)n-r
P(x = r) = 4C(1/5)r(4/5)4-r for r = 0, 1, 2, 3, 4
Hence, the probability distribution is given as:
X
0
1
2
3
4
P(X)
256/625
256/625
96/625
16/625
1/625

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