NCERT Solutions for Class 10 Maths Exercise 15.2 Chapter 15 Probability- FREE PDF Download
NCERT Class 10 Maths Ch 15 is one of the most important ones in the NCERT syllabus. Duly following NCERT Solutions for Class 10 Maths
Chapter 15 ensures you that there will be no hindrance when you opt for more advanced branches of Maths. This is where CoolGyan comes in. Our free Class 10 Probability solutions will help you understand the chapter thoroughly.
NCERT Solutions for Class 10 Maths Chapter 15 – Probability
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?
(T, T) (T, W) (T, TH) (T, F) (T, S)
(W, T) (W, W) (W, TH) (W, F) (W, S)
(TH, T) (TH, W) (TH, TH) (TH, F) (TH, S)
(F, T) (F, W) (F, TH) (F, F) (F, S)
(S, T) (S, W) (S, TH) (S, F) (S< S)
where T = Tuesday, W = Wednesday, Th = Thursday, F = Friday, S = Saturday
Total number of favourable outcomes = 5 x 5 = 25
(i) The favourable outcomes of visiting on the same day are (T, T), (W, W), (TH, TH), (F, F) and (S, S).
Number of favourable outcomes = 5
Hence required probability = 
(ii) The favourable outcomes of visiting on consecutive days are (T, W), (W, T), (W, TH), (TH, W), (TH, F), (F, TH), (S, F) and (F, S).
Number of favourable outcomes = 8
Hence required probability = 
(iii) Number of favourable outcomes of visiting on different days are 25 – 5 = 20
Number of favourable outcomes = 20
Hence required probability = 
2. A die is numbered in such a way that its faces show the numbers 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:
(i) even
(ii) 6
(iii) at least 6?
It is clear that total number of favourable outcomes = 
= 36
(i) Even scores are: 2, 4, 4, 4, 4, 8, 4, 4, 8, 4, 6, 4, 6, 6, 8, 8, 12
Number of favourable outcomes of getting total score even are 18
Hence P (getting total score even) = 
(ii) Number of favourable outcomes of getting total score 6 are 4
Hence P (getting total score 6) = 
(iii) Total score at least 6 = 7, 8, 8, 6, 6, 9, 6, 6, 9, 7, 8, 8, 9, 9, 12
Number of favourable outcomes of getting total score at least 6 are 15
Hence P (getting total score at least 6) = 
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.
blue balls in the bag. Total number of balls in the bag = 
Now, 
= Probability of drawing a blue ball = 
And 
= Probability of drawing a red ball = 
But according to question, 
= 
= 2
Hence, there are 10 blue balls in the bag.
4. A box contains 12 balls out of which 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 of what it was before. Find 
Therefore, total number of favourbale outcomes = 12
The number of favourable outcomes (Black balls) =
Therefore = P (getting a black ball) = 
If 6 more balls put in the box, then
Total number of favourable outcomes = 12 + 6 = 18
And Number of favourable outcomes = 
P2 = P (getting a black ball) = 
According to question, 
= 
⇒⇒ x+618=x6x+618=x6
⇒⇒ 6x+36=18x6x+36=18x
18x−6x=3618x−6x=36
⇒⇒ 12x=3612x=36
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 
Find the number if blue balls in the jar.
Let there be
green marbles.Therefore, Favourable number of outcomes =
P(Green ball) = 
But P(Green ball) =
= 16Therefore, number of green marbles are 16
And number of blue marbles = 24 – 16 = 8
