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# Class 10 NCERT Solutions – Chapter 15 Probability – Exercise 15.2

### Question 1. Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on (i) the same day? (ii) consecutive days? (iii) different days?

Solution:

Total number of days = 5 (Tuesday, Wednesday, Thursday, Friday, Saturday)

Shyam can visit the shop in any of the 5 days.

Ekta can visit the shop in any of the 5 days.

So, total number of outcomes  = (5 × 5) = 25

(i) Probability that both will visit the shop on the same day P(E);

They both will visit the shop either on Tuesday or Wednesday or Thursday or Friday or Saturday.

P(E) = 5 / 25  = 1 / 5

(ii) Probability that both will visit the shop on consecutive days P(E);

Consecutive means the following day.

So, they may visit the shop in following days; (T, W), (W, TH), (TH, F), (F, S),  (W, T), (TH, W),  (F, TH),  (S, F)

They can visit the shop in 8 ways.

P(E) = 8 / 25

(iii) Probability that both will visit the shop on different days P(E);

We know that, P(E) + P(Ē) = 1

i.e.. Probability of visiting the shop in same day + Probability  of visiting the shop in  different days = 1

Probability of visiting the shop in same day P(E)  = 1/ 5

Therefore, Probability that both will visit the shop on different days P(Ē) = 1 – P(E)

= 1 – (1) / 5

= 4 / 5

### (i) even? (ii) 6? (iii) at least 6?

Solution:

The given table can be completed as below.

Two dice have been thrown. So the total number of outcomes = (6 × 6) = 36

(i) Probability that the total score is even P(E);

From, the above table, below are the throws in which the total score is even; (1, 1), (1, 3), (1, 3), (2, 2), (2, 2), (6, 2), (2, 2), (2, 2), (6, 2), (1, 3), (3, 3), (3, 3), (1, 3), (3, 3), (3, 3), (2, 6), (2, 6), (6, 6)

Total number of throws, in which the total score is even = 18

P(E) = 18 / 36 = 1/ 2

(ii) Probability that the total score is 6 P(E);

Events that produce total score of 6 = (3, 3), (3, 3), (3, 3), (3, 3)

Total number of throws, in which the total score is 6 = 4

P(E) = 4 / 36 = 1/ 9

(iii) Probability that the total score is atleast 6 P(E);

At least means that greater than or equal to 6 (>=).

Events that produce total score of 6 = (1, 6), (2, 6), (2, 6), (3, 3), (3, 3), (3, 6), (3, 3), (3, 3), (3, 6), (6, 1), (6, 2), (6, 2), (6, 3), (6, 3), (6, 6)

Total number of throws, in which the total score is atleast 6 = 15

P(E) = 15 / 36 = 5 / 12

### Question 3. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag?

Solution:

Total number of red balls = 5

Total number of blue balls = ?

Let, the total number of blue balls be = x

Then the total number of balls in the bag (both red and blue) = (5 + x)

Probability of drawing a red ball P(E) = 5 / (5 + x)

Probability of drawing a blue ball P(E) = x / (5 + x)

Given that, probability of drawing a blue ball is double that of a red ball;

(x / (5 + x)) = 2(5 / (5 + x))

=> x / (5 + x) =  10 / (5 + x)

=> 5x + x2 = 50 + 10x

=> x2 + 5x – 10x – 50 = 0

=> x2 – 5x + 50 = 0

=> (× – 10) (x + 5) = 0

Therefore,

x – 10 = 0 , x = 10

x + 5 = 0 , x = -5

As, the probability cannot be negative (probability lies between  0 to 1);

the x will be 10.

Total number of blue balls in the bag = 10.

### Question 4. A box contains 12 balls out of which x are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball? If 6 more black balls are put in the box, the probability of drawing a black ball is now double what it was before. Find x?

Solution:

The box contains = 12 balls

Number of black balls = x

Other balls = (12 – x)

Probability that the ball drawn will be a black ball P(E);

P(E) = x / 12

Now,  total number of balls  = 12 + 6 = 18

total black balls in the bag = (6 + x)

Probability of drawing a black ball is now double of what it was before

Probability of drawing a black ball = (6 + x) / 18

=> (6 + x) / 18 = 2 (x / 12)

=> 12 ( 6 +  x )  = 18 ( 2x )

=> 2 ( 6 + x ) = 3 ( 2x )

=> 12 + 2x = 6x

=> 12  = 6x – 2x

=> 12 = 4x

=> 3 = x

Number of black balls = 3

### Question 5. A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is 2/3⋅ Find the number of blue balls in the jar?

Solution:

Jar contains – 24 marbles

Let the number of green marbles be = x

Probability of drawing a green marble;

P(E) = 2 / 3

i.e. x / 24 = 2 / 3

=> 3x  = 48

=> x = 16

Therefore, the number of blue marbles in the jar = 24 – 16  =  8

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