These NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.2 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 10 Maths 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:

The day are given tues, wed, thr, fri, sat Sample space of the days.

Total number of sample space = 5 x 5

n(S) = 25

(i) Let E be the event on the same day

E = (t, t), (w, w), (th, th), (f, f), (s, s)

Number of events E occured the same day = 5

P(E) = \(\frac { 5 }{ 25 }\) = \(\frac { 1 }{ 2 }\) [P(E) = \(\frac { n(E) }{ n(S) }\)]

(ii) Let F be the event on the consecutive days

n( F) = 8

n(S) = 25

P(F) = \(\frac { n(E) }{ n(S) }\) = \(\frac { 8 }{ 25 }\)

(iii) Let G be the event of different days

n(G) = 20

P(G) = \(\frac { 20 }{ 25 }\) = \(\frac { 4 }{ 25 }\)

Question 2.

A die is numbered in such a way that its faces show the number 1, 2, 2, 3, 3, 6. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score on the two throws:

What is the probability that the total score is at least 6?

(i) even

(ii) 6

(iii) at least 6

Solution:

Question 3.

A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is doubles that of a red ball, determine the number of blue balls in the bag.

Solution:

Let the number of blue balls be x

Given that the number of red balls is 5

The total number of balls in bag = x + 5

P (getting a red bails) = \(\frac { x+5 }{ 5 }\)

P (getting a blue ball) = \(\frac { 5 }{ x+5 }\)

According to question

2(\(\frac { x+5 }{ 5 }\)) = \(\frac { 5 }{ (x+5) }\)

10 (x + 15) = x² + 5x

x² – 5x – 50 = 0

x² – 10x + 5x – 50 = 0

x[x – 10] – 5 (x – 10] = 0

(x – 10) (x + 5) = 0

x = – 5 which is not possible.

x = 10

Hence, the number of blue balls is 10.

Question 4.

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

Solution:

The total number of balls containing in a box = 12

The total number of black bails = x ;

P (getting a black ball) = \(\frac { x }{ 12 }\)

Now,

When 6 more black balls are put in the box then

The total number of balls in the box = 12 + 6 = 8

and The total number of black balls = x + 6

P (getting a black ball after putting 6 more balls) = \(\frac { x+6 }{ 18 }\)

According to question

2(\(\frac { x }{ 12 }\)) = \(\frac { x+6 }{ 18 }\)

36x = 12x + 72

24x = 72

x = 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 \(\frac { 2 }{ 3 }\). Find the number of blue balls in the jar.

Solution:

The total number of marbles in the jar = 24

Let the total number of green marbles in the jar be x

Then the total number of blue marbles = 24 – x

P (getting a green marble) = \(\frac { x }{ 24 }\)

According to question

P (getting a green marble) = \(\frac { 2 }{ 3 }\)

\(\frac { x }{ 24 }\) = \(\frac { 2 }{ 3 }\)

x = \(\frac { 24×2 }{ 3 }\)

The total number of blue marbles = 34 – x

= 24 – 16 = 8

Hence, the jar contains 8 blue marbles.