## NCERT Solutions Class 10 |

Sum on 2 die | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |

Probability | 1/36 | 1/18 | 1/12 | 1/9 | 5/36 | 1/6 | 5/36 | 1/9 | 1/12 | 1/18 | 1/36 |

**(ii)** As shown above there are a total of 36 possible outcomes for this event. Hence, the statement made by the student is incorrect.

**23. A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.**

**SOLUTION**

Possible outcomes on 3 coin tosses :

HHH, HHT, HTH, HTT, THH, THT, TTH, TTT

Total outcomes = 8

Hanif will win the game if the outcome is HHH or TTT.

P(Hanif will lose) = 1 - P(Hanif will win)

= 1 - 2/8 = ¾

Hence, the probability that hanif will lose the game is 3/4.

**24. A die is thrown twice. What is the probability that**

**(i) 5 will not come up either time? (ii) 5 will come up at least once?**

**[Hint : Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]**

**SOLUTION**

Possible outcomes on 2 dice throws:

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

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

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

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

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

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

Total possible outcomes = 36

**(i)** Number of outcomes without 5 = 25

P(5 does not come up either time) = 25/36

**(ii)** P(5 comes up at least once) = 1 - P(5 does not come up either time)

= 1 - 25/36

= 11/36

**25. Which of the following arguments are correct and which are not correct? Give reasons for your answer.**

- If two coins are tossed simultaneously there are three possible outcomes?two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is 1/3.
- If a die is thrown, there are two possible outcomes?an odd number or an even number. Therefore, the probability of getting an odd number is 1/2.

**SOLUTION**

**I.** Possible outcomes when 2 coins are tossed:

(H, H), (H,T), (T,H), (T,T)

Total outcomes = 4

Number of outcomes with both head and tails = 2

P(getting both Heads and Tails) = 2/4 = 1/2

Therefore, the given statement is incorrect.

**II.** Possible outcomes when a die is thrown:

1, 2, 3, 4, 5, 6

Total outcomes = 6

Number of outcomes with an odd number = 3

P(getting an odd number) = 3/6 = 1/2

Therefore, the given statement is true.

**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**

Possible outcomes when Shyam and Ekta visit the shop in the same week:

(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)

[T = Tuesday, W = Wednesday, Th = Thursday, F = Friday, and S = Saturday]

Total outcomes = 25

**I.** Number of outcomes where both visit on the same day = 5

P(they visit the shop on the same day) = 5/25 = 1/5

**II.** Number of outcomes where both visit on consecutive days = 8

P(they visit the shop on consecutive days) = 8/25

**III.** Number of outcomes where both visit on different days = 20

P(they visit the shop on different days) = 20/25 = 4/5

**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?**

**SOLUTION**

Completed table will be:

+ | 1 | 2 | 2 | 3 | 3 | 6 |
---|---|---|---|---|---|---|

1 | 2 | 3 | 3 | 4 | 4 | 7 |

2 | 3 | 4 | 4 | 5 | 5 | 8 |

2 | 3 | 4 | 4 | 5 | 5 | 8 |

3 | 4 | 5 | 5 | 6 | 6 | 9 |

3 | 4 | 5 | 5 | 6 | 6 | 9 |

6 | 7 | 8 | 8 | 9 | 9 | 12 |

Total outcomes = 36

**(i)** Number of outcomes with even score = 18

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

**(ii)** Number of outcomes with score of 6 = 4

P(getting score of 6) = 4/36 = 1/9

**(iii)** Number of outcomes with score of at least 6 = 9

P(getting a score of at least 6) = 9/36 = 1/4

**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**

Let the number of blue balls in the bag be x

Total number of balls in the bag = 5 + x

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

5/(5 + x) × 2 = x/(5 + x)

10 = x

Therefore, the bag contains 10 blue balls.

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

**SOLUTION**

Total number of balls in the box = 12

Number of black balls = x

P(drawing a black ball) = x/12

Total number of balls in the box when 6 black balls are added = 12 + 6 = 18

Number of black balls when 6 are added = x + 6

P(drawing a black ball) × 2 = P(drawing a black ball after adding 6 black balls)

x/12 × 2 = (x + 6)/18

3x = x + 6

2x = 6

x = 3

Hence, there were originally 3 black balls in the bag.

**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**

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

Then the number blue marbles = 24 - x

Total number of marbles in the jar = 24

P(drawing a green marble) = 2/3

x/24 = 2/3

x = 16

Number of green marbles = 16

Hence, number of blue marbles = 24 - 16 = 8

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