**NCERT Solutions for Class 10 Maths Exercise 15.2 of Chapter 15 Probability**

NCERT Solutions for Class 10 Maths Exercise 15.2 of Chapter 15 Probability gives you an idea of solving the questions based on the probability in the CBSE class 10 board maths question paper and other competitive exams maths paper. All the NCERT solutions are created by a qualified maths teacher who has huge experience in teaching mathematics to classes of 9 ,10,11, and 12. These NCERT solutions are solved as per the CBSE norms and updated syllabus of class 10 as prescribed by CBSE.

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**NCERT Solutions for Class 10 Maths of Chapter 15 Probability**

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**NCERT Solutions for Class 10 Maths Exercise 15.2 of Chapter 15 Probability**

**Q1.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?**

Ans. We are given that Shyam and Ekta visits a particular shop in the same week from Tuesday to Saturday

Let Shyam visits on T,W,Th,F,S and Ekta visits on T,W,Th,F,S

The possible combination of the days they visit the shop in a week are **(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)**,

(i) The outcomes which represents their visits on the same days are (TT),(WW),(Th Th),(FF) and (SS),therefore number of favourable outcomes = 5

Total outcomes are = 25

P(E) = Favourable outcomes/Total possible outcomes

P(Visit on the same day) = 5/25 = 1/5

(ii) The outcomes which represents their visits on consecutive days are (T,W),(W,T),(Th,F),(F,S), number of favourable outcomes = 4

Total possible outcomes = 25

P(Shyam visit the shop on consecutive days) = 4/25

P(Ekta visit the shop on consecutive days) = 4/25

P(Both visit the shop on consecutive days) = 2×(4/25) = 8/25

(iii) The outcomes which represents their visits on different days are (T,W),(T,Th),(T,F),(T,S),(W,T),(W,Th),(W,F),(W,S),(Th,T),(Th,W),,(Th,F),(Th,S),(F,T),(F,W),(F,Th),,(F,S),(S,T),(S,W),(S,Th),(S,F), therefore number of favourable outcomes are =20

P( Their visit on different days) = 20/25 = 4/5

**NCERT Solutions for Class 10 Maths Exercise 15.2 of Chapter 15 Probability**

**Q2.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 probablity that the total score is

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

Ans.We are given that the dies whose faces show the numbers 1,2,2,3,3,6 are thrown two times,so the outcomes are as follows

(1,1),(1,2),1,2),(1,3),(1,3),(1,6),(2,1),(2,2),2,2),(2,3),(2,3),(2,6),(2,1),(2,2),(2,2),(2,3),(2,3),(2,6)(3,1),(3,2),3,2),(3,3),(3,3),(3,6),(3,1),(3,2),3,2),(3,3),(3,3),(3,6),(6,1),(6,2),6,2),(6,3),(6,3),(6,6),

Total scores in both dies are

(2,3,3,4,4,7),(3,4,4,5,5,8),(3,4,4,5,5,8),(4,5,5,6,6,9),(4,5,5,6,6,9),(7,8,8,9,9,12)

Therefore the complete score tables will be as follows

(i) The outcomes which represent total score even are 2,4,4,4,4,8,4,4,8,4,6,6,4,6,6,8,8,12,therefore favourable outcomes are = 18

Total possible outcomes are = 36

P(score is even) = 18/36 = 1/2

(ii) The outcomes which represent total score 6 are 6,6,6,6, therefore favourable outcomes are = 4

Total possible outcomes are = 36

P(score is 6) = 4/36 = 1/9

(iii)The outcomes which represent total score at least 6 are 7,8,8 6,6,9,6,6,9,7,8,8,9,9,12,,therefore favourable outcomes are = 15

P(at least 6) = 15/36 = 5/12

**NCERT Solutions for Class 10 Maths Exercise 15.2 of Chapter 15 Probability**

**Q3.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.**

Ans. Let the number of blue balls in the bag are = x

The number of given red balls in the bag are = 5

Total balls in the bag are = (5 +x)

P(drawing a blue ball) = (number of blue balls)/Total balls = x/ (5 +x)

P(drawing a red ball) = (number of red balls)/Total balls = 5/ (5 +x)

According to question

P(drawing a blue ball) = 2×P(drawing a red ball)

x/ (5 +x) = 2× 5/ (5 +x)

x/ (5 +x) = 10/ (5 +x)

x = 10

**Therefore number of blue balls in the bag are 10**

**NCERT Solutions for Class 10 Maths Exercise 15.2 of Chapter 15 Probability**

**Q4.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 of what it was before. Find x**

Ans.We are given the number of balls in the bag are = 12

And the number of black balls in the bag are = x

∴P( drawing a black ball) = (number of black balls in the bag)/(Total number of black balls in the bag) = x/12

Since 6 more black balls are put in the box therefore total number of balls in the bag are =12 +6 =18

∴The number of black balls in the bag becomes = (x + 6)

Now,the probability of drawing a black ball is = (x+6)/18

According to question

The probability of drawing a black ball in the second case = 2×The probability of drawing a black ball in first case

(x+6)/18 = 2× x/12

12( x+ 6) = 2x× 18

12x + 72 = 36x

24x = 72

x = 3

**NCERT Solutions for Class 10 Maths Exercise 15.2 of Chapter 15 Probability**

**Q5.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.**

Ans.Let the number of blue balls in the jar are = x

Total number of marbles in the jar are = 24

The number of green marbles in the jar = 24 – x

∴ P( drawing a green marbles) = (number of green marbles)/(Total marbles in the jar)

=(24-x)/24

It is given to us that probability of drawing a green marbles is = 2/3

According to question,we have

(24-x)/24 = 2/3

72 – 3x = 48

3x = 24

x = 8

Therefore the number of blue balls in the jar are 8

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