Here we provide Sharma Class 12 Ex 31.6 Solutions Chapter 31 Probability for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 31.6 Solutions Chapter 31 Probability book pdf download. Now you will get step-by-step solutions to each question.

Textbook | NCERT |

Class | Class 12th |

Subject | Maths |

Chapter | 31 |

Exercise | 31.6 |

Category | RD Sharma Solutions |

**RD Sharma Class 12 Ex 31.6 Solutions Chapter 31 Probability**

### Question 1. A bag A contains 5 white and 6 black balls. Another bag B contains 4 white balls and 3 black balls. A ball is transferred from bag A to bag B .then a ball is drawn out from bag B. find out probability that the ball drawn being black?

**Solution:**

Bag ‘A’ contains 5 white balls and 6 black balls

Bag ‘B’ contains 4 white balls and 3 black balls

There are two ways of transferring ball:

1. Transfer 1 white ball from A to B and then draw a black ball from bag B

2. Transfer 1 black ball from A to B and then draw a black ball from bag B

Let E1, E2, A be events as follows:

E1: White ball drawn from bag 1

E2: Black ball drawn from bag 2

A: PROBABILITY OF BALL BEING BLACK

Step 1: Remove probability of drawing 1 white ball from bag AEVENT 1(E1): One white ball drawn from bag A

P(E1) = 5/11 -(1)

Step 2: Remove probability of drawing 1 black ball from bag AEVENT 2(E2): One black ball drawn from bag A

P(E2) = 6/11 -(2)

Step 3: Remove the probability that a ball drawn from bag B is black

with 1 white ball increased and 1 black ball increased(A): Drawing one black ball from bag B

P(B/E1) = 3/8 -(E1 increased 1 white ball in bag B)(3)

P(B/E2) = 4/8 -(E2 has increased 1 black ball in bag B)(4)

Step 4: Apply the formula and put the removed values of eventsBy the law of probability

P(A) = P(E1)*P(A/E1)+P(E2)*P(A/E2)

= (

5/11)*(3/8)+(6/11)*(4/8) -(From 1, 2, 3, 4)= 39/88

The probability of ball being black is 39/88

### Question 2: A Purse contains 2 silver coins and 4 copper coins. Another purse contains 4 silver and 3 copper coins. If a coin is pulled from a random of 2 purses find the probability of the coin being silver?

**Solution:**

Purse ‘1’ contains 2 silver coins and 4 copper coins

Purse ‘2’ contains 4 silver coins and 3 copper coins

One coin is from one of the purses and the coin is silver

Step 1: Finding probability of selecting purse 1E1: Selecting purse 1

P(E1) = 1/2

Step 2: Finding probability of selecting purse 2E2: Selecting purse 2

P(E2) = 1/2

A: Probability of drawing silver coin

P(A|E1) = 2/6

= 1/3 -(drawing silver coin from purse 1)

P(A|E2) = 4/7 -(drawing coin from purse 2)

From the law of probability

P(A) = P(E1)*P(A|E1)+P(E2)*P(A|E2)

= (1/2)*(1/3)+(1/2)*(4/7)

= (1/6)+(4/14)

= 19/42

The probability of drawing silver coin from two purses is 19/42

### Question 3: One bag contains 4 yellow and 5 red balls and another bag contains 6 yellow and 3 red balls. A ball is transferred from bag 1 to bag 2. Find the probability of the ball being yellow if drawn from bag 2?

**Solution:**

Bag 1 contains 4 yellow and 5 red balls

Bag 2 contains 6 yellow and 3 red balls

A ball is transferred from bag1 to bag 2 this can be done in 2 ways:

1. Transferring yellow ball to bag 2

2. Transferring red ball to bag 2

Let E1, E2, and A be event such as:

E1: One yellow ball drawn from bag 1

E2: One red ball drawn from bag 2

A: One yellow drawn from bag 2

Step 1: Find the probability of event E1P(E1) = 4/9 -(1)

Step 2: Find the probability of event E2P(E2) = 5/9 -(2)

Step 3: Find the probability of drawing yellow ball from

bag 2 after transfer of balls from bag 1P(A|E1) = 7/10 -(3) (Since E1 has increased 1 yellow ball in bag 2)

P(A|E2) = 6/10 -(4) (Since E2 has increased 1 red ball in bag 2)

Step 4: Applying the law of probabilityP(A) = P(E1)*P(A|E1)+P(E2)*P(A|E2)

= (

4/9)*(7/10)+(5/9)*(6/10) -(From 1, 2, 3, 4)= 58/9

= 29/45

The probability of drawing yellow ball is 29/45

### Question 4: A bag contains 3 white balls and 2 black balls. Another bag 2 contains 2 white balls and 4 black balls. One bag is chosen at random and a ball is drawn from that bag. Find the probability of the ball being white?

**Solution:**

Bag 1 contains 3 white balls and 2 black balls

Bag 2 contains 2 white balls and 4 black balls

One bag is chosen at random

Let E1, E2, and A be events such as

E1: Selecting bag I

E2: Selecting bag II

A: Drawing white ball

Step 1: Probability of selecting bag 1P(E1) = 1/2 -(1)

Step 2: Probability of selecting bag 2P(E2) = 1/2 -(2)

Step 3: Find the probability of drawing white ballP(A|E1) = P (Drawing white ball from bag 1)

= 3/5 -(iii)

P(A|E2) = P (Drawing white ball from bag 2)

= 2/6 = 1/3 (iv)

Step 4: Use the law of probabilityP(A) = P(E1)*P(A|E1)+P(E2)*P(A|E2) -(From 1, 2, 3, 4)

= (1/2)*(3/5)+(1/2)*(1/3)

= 7/15

The probability of ball being white is 7/15

**Question 5: The contents of 3 bags are as follows**

**Bag I: 1 white, 2 black, 3 red balls**

**Bag II: 2 White, 1 black, 1 Red balls**

**Bag III: 4 White, 5 black, 3 Red balls**

**A bag is selected at random and two balls are drawn, what is **the **probability that balls are red and white?**

**Solution:**

Let E1, E2, E3, A be the events such as

E1: Selecting BAG I

E2: Selecting BAG II

E3: Selecting BAG III

A: Drawing a red and a white ball

Step 1: Finding the probability of selecting one bag at randomP(E1): 1/3 -(1)

P(E2): 1/3 -(2)

P(E3): 1/3 -(3)

Step 2: Finding the probability of drawing one red and one white ballP(A|E1) = P (Drawing one red and one white ball from bag I)

=

^{1}C_{1 * }^{3}C_{1}/^{6}C_{2 }(^{n}C_{r}= n!/r!*(n-r)!)_{ }_{ }

= (2 * 1) / (4 * 3) /2= 1/5 -(4)

P(A|E2) = P(Drawing ball 1 red and white ball from bag II)

= (2*1)/4*3/2

= 1/3 -(5)

P(A|E3) = P(Drawing ball 1 red and white ball from bag III)

= 2/11 -(6)

Step 3: Applying the law of probabilityP(A)= P(E1)*P(A|E1)+P(E2)*P(A|E2)+P(E3)*P(A|E3) -(From 1, 2, 3, 4, 5, 6)

= (1/3*1/5)+(1/3*1/3)+(1/3*2/11)

= (1/15)+(1/9)+(2/33)

= 118/495

The required probability is 118/495

### Question 6: A unbiased coin is tossed is the result is head a pair of unbiased dice is rolled and the sum of the numbers is obtained is noted. If the result is 12 then a card from a pack of 11 cards numbered 2,3,4,5……….12 is picked and a number of cards are noted. What is the probability that the number is 7 or 8?

**Solution:**

An unbiased coin is tossed then:

1. If the head occurs a pair of dice is rolled and the sum of them is either 7 or 8

2. If

tail occurs a card is drawn from card numbered 2,3,4,5 ….. 12. And is 7 or 8Let E1, E2, & A be events as

E1: A head occurs

E2: A tail occurs

A: The noted number is 7 or 8

Step 1: Find the probability of tail comes up in a single coinP(E1) = 1/2 -(1)

Step 2: Find the probability of head comes up in a single coinP(E2) = 1/2 -(2)

Step 3: Probability that number noted is 7 or 8P(A|E1) = P[Pair of dice shows 7 or 8 as sum]

(Sum of pair is 7 or 8 when – (1,6), (2,5), (3,4),

(4,3), (5,2), (6,1), (6,2), (5,3), (4,4), (3,5), (2,6))

P(A|E1) =11/36 -(3)

P(A|E2) = P[7 or 8 occurs on cards drawn numbered 2,3,4….12]

= 2/11 -(4)

Step 4: Applying the law of probabilityP(A) = P(E1)*P(A|E1)+P(E2)*P(A|E2)

= (1/2)*(11/36)+(1/2)*(2/11)

=193/792

The required probability is 193/792

### Question 7. A factory has two machines A and B. The past records show that machine A produce the 60% of the item of output and B produce 40%. Further, 2% of items produced by machine A were defective and 1% of B were defective. If item is drawn at random what is the probability that it is defective?

**Solution:**

Let E1, E2, A be the events such as

E1: Item produced by machine A

E2: Item produced by machine B

A: Product is defective

Step 1: Remove the probability of E1 and E2P(E1) = 60%

= 60/100 -(1)

P(E2) = 40%

= 40/100 -(2)

Step 2: Find the probability that the product is defectiveP(A|E1) = P(Defective item from machine A)

= 2%

= 2/100 -(3)

P(A|E2) = P(Defective item from machine B)

= 1%

1/100 -(4)

Step 3: Applying the law of probabilityP(A) = P(E1)*P(A|E1)+P(E2)*P(A|E2)

= (60/100)*(2/100)+(40/100)+(1/100) (From 1, 2, 3, 4)

= 160/10000

= 0.016

The required probability is 0.016

### Question 8. A bag A contains 8 white and 7 black balls and bag B contains 5 white and 4 black balls. One ball is picked from bag A and mixed up with balls in bag B. A ball is drawn at random from bag B. the probability that the ball is white?

**Solution:**

Transfer if the ball from bag A to bag B can be done in two ways:

1. One white ball is transferred from bag A to B, then a white ball is drawn

2. One black ball is transferred from bag A to B, then a black ball is drawn.

Let E1, E2 & A be the events as:

E1: One white ball from bag A

E2: One black ball from bag A

A: One white ball from bag B

Step 1: Find the probability of balls drawn from bag AP(E1) = 8/15 -(1)

P(E2) = 7/15 -(2)

Step 2: Find the probability of a ball drawn from bag BP(A|E1) = 6/10 -(3) (Since E1 has increase white ball by 1 in bag B)

P(A|E2) = 4/10 -(4) (E2 increase black ball in bag 2)

Step 3: Applying the law of probabilityP(A) = P(E1)*P(A|E1)+P(E2)*P(A|E2)

= (8/15)*(6/10)+(7/15)*(4/10) -(From 1, 2, 3, 4)

= 83/150

The required probability is 83/150

### Question 9. A bag 1 contains 4 white and 5 black balls another bag contains 3 white and 4 black balls. A ball is taken out without seeing color and put it in another bag. A ball is taken out from it later. Find the probability that the ball is white?

**Solution:**

A ball is taken out from bag I and put in bag II without seeing its color.

Then the ball is taken out from bag II and it is white

Step 1: Find the probability that the ball is drawn from bag 1 is whiteP(W1) = 4/9 -(1)

Step 2: Find the probability that the ball is drawn from bag 1 is blackP(B1) = 5/9 -(2)

Step 3: Find the probability after all is moved to bag 2P(W2/B1) = P (White ball drawn from bag 2 after B1 transfer)

= (3/8) -(3)

P(W2/W1) = P(White ball drawn from bag 2 after W1 transfer)

= (4/8)

= (1/2) -(4)

Step 4: Applying the law of probabilityP(White ball from bag 2)= P(B1)*P(W2/B1)+P(W1)*P(W2/W1)

= (5/9)*(3/8)+(4/9)*(1/2)

= 31/72

The required probability 31/72

### Question 10. One bag contains 4 white & 5 black balls another bag contains 6 white & 7 black balls. A ball is transferred from bag I to bag II and the ball is drawn from bag II. Find the probability that the ball is white?

**Solution:**

A ball is taken from bag 1 and without seeing the color the ball is mixed with balls in bag 2. Then the ball is drawn from the bag 2, what is the probability that the ball is white

Step 1: Find the probability that the ball removed from bag 1 is whiteP(1 White ball from bag 1) = P(W1)

= 4/9 -(1)

Step 2: Find the probability that the ball removed from bag 1 is blackP(1 Black ball drawn from bag 2) = P(B1)

= 5/9 -(2)

Step 3: Find the probability that the ball drawn from bag 2 is white after transferP(White ball transferred from bag 1 to bag 2) = P(W2/W1)

= 7/14

= 1/2 -(3)

P(Black ball transferred from bag 1 to bag 2) = P(W2/B1)

= 6/14

= 3/7 -(4)

Step 4: Applying the law of probabilityP(1 WHITE FROM BAG II) = P(W1)*P(W2/W1)+P(B1)*P(W2/B1)

= (4/9)*(1/2)+(5/9)*(3/7) (From 1, 2, 3, 4)

= 58/126

= 29/63

The required probability 29/63

### Question 11. An urn contains 10 white balls and 3 black balls. Another urn contains 3 white balls and 5 black balls. Two balls are drawn from urn I and put in the second urn and then a ball is drawn later. Find the probability that ball is white?** **

**Solution:**

**URN “1”**

10 white balls (10 WB)3 black balls (3 BB) |

**URN “2”**

3 white balls (3 WB)5 black balls (5 BB) |

Let U1_{2W, }U1_{1W1B, }U1_{ 2B }be the events of transferring 2 white balls (2W), one white & black ball(1W1B), and transferring 2 black balls 2BB from urn I to urn II.

P(drawing two white balls from urn I) = ^{10}C_{2}^{ }/^{13}C_{2 }= 45/78 -(1)

P(drawing 1 white and 1 black ball from urn I) = ^{10}C_{1 }^{3}C_{1}/^{13}C_{1 }= 30/78 -(2)

P(drawing 2 black balls from urn I) = ^{3}C_{2}/^{13}C_{2 }=_{ }3/78 -(3)

Let U2(w) drawing from the white ball from urn II. There will be three cases based on above events

I | II | III | |

5 WHITE | 4 WHITE | 3 WHITE | |

5 BLACK | 6 BLACK | 7 BLACK | |

TOTAL | 10 | 10 | 10 |

P(Based on event 1BASED ON EVENT 1) = ^{5}C_{1}/^{10}C_{1}

= 5/10

= 1/2 -(4)

P(Based on event 2) = ^{4}C_{1}/^{10}C_{1}

= 4/10

= 2/5 -(5)

P(Based on event 3) = ^{3}C_{1}/^{10}C_{1}

_{= }3/10 -(6)_{ }

From the law of probability

P(U_{2W}) = (45/78)*(1/2)+(30/78)*(2/5)+(3/78)*(3/10) -(From 1, 2, 3, 4, 5, 6)

= 59/130

The required probability 59/130

### Question 12. A bag contains 6 red and 8 black balls another bag contains 8 red and 6 black balls. A ball is drawn from the bag I and without seeing color is kept in bag II. Then the ball is drawn from bag II find the probability that ball is red?

**Solution:**

Given:Bag I contain 6 red(R1) and 8 black(B1) ballsBag II contains 8 red(r2) and 6 black(B2) balls

A ball is drawn from bag I and without seeing color is kept in bag II,

then the ball drawn from bag 2 is Red

P(One red ball from bag 2)

= P((B1 n R2)U (R1 n R2))

= P(B1 n R2) + P(R1 n R2)

= P(B1)*P(B1/R2)+P(R1)*P(R1/R2)

= (8/14)*(8/15)+(6/14)*(9/15)

= 118/210

= 59/105

The required probability is 59/105

### Question 13. Three machines E1, E2, E3 produce 50%,25%, and 25% respectively of the daily total output of electric bulbs. It is known that 4% of each tube produced one each of machine E1 & E2 is defective and 5 % of those produced on E3 is defective. If one tube is picked from a day’s production, calculate the probability that the tube is defective?

**Solution:**

Let D be the event that the bulb is defective

A1, A2 & A3 be the events that tube is produced from the machine E1, E2, & E3

P(D) = P(A

_{1})*P(D|A_{1})+P(A_{2})*P(D|A_{2}) -(1)P(A1) = 50/100

= 1/2 -(2)

P(A2) = 25/100

= 1/4 -(3)

P(A3) = 25/100

= 1/4 -(4)

P(D|A1) = 4/100 = P(D|A2)

= 1/25 -(5)

P(D|A3) = 5/100

= 1/20 -(6)

By applying the law of probability:

P(D) = (1/2)*(1/25) + (1/4)*(1/25) + (1/4)*(1/25)

= 17/400

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.