  # Important Questions for CBSE Class 10 Maths Chapter 15 - Probability 3 Mark Question

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CBSE Class 10 Maths Chapter-15 Probability Important Questions

## 3 Mark Questions

1. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see figure) and these are equally likely outcomes. What is the probability that it will point at:
(i) 8?
(ii) an odd number?
(iii) a number greater than 2?
(iv) a number less than 9? Ans. Out of 8 numbers, an arrow can point any of the numbers in 8 ways. Total number of favourable outcomes = 8
(i) Favourable number of outcomes = 1
Hence, P (arrow points at 8) = (ii) Favourable number of outcomes = 4
Hence, P (arrow points at an odd number) = (iii) Favourable number of outcomes = 6
Hence, P (arrow points at a number > 2) = (iv) Favourable number of outcomes = 8
Hence, P (arrow points at a number < 9) = =1

2. A dice is thrown once. Find the probability of getting:
(i) a prime number.
(ii) a number lying between 2 and 6.
(iii) an odd number.
Ans. Total number of favourable outcomes of throwing a dice = 6
(i) On a dice, the prime numbers are 2, 3 and 5.
Therefore, favourable outcomes = 3
Hence P (getting a prime number) = (ii) On a dice, the number lying between 2 and 6 are 3, 4, 5.
Therefore, favourable outcomes = 3
Hence P (getting a number lying between 2 and 6) = (iii) On a dice, the odd numbers are 1, 3 and 5.
Therefore, favourable outcomes = 3
Hence P (getting an odd number) = 3. 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.
Ans. The outcomes associated with the experiment in which a coin is tossed thrice:
HHH, HHT, HTH, THH, TTH, HTT, THT, TTT
Therefore, Total number of favourable outcomes = 8
Number of favourable outcomes = 6
Hence required probability = 4. 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?
Ans. Complete table is as under: It is clear that total number of favourable outcomes = 6 x 6 = 36
(i) 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) Number of favourable outcomes of getting total score at least 6 are 15
Hence P (getting total score at least 6) = 5. 18 cards numbered 1, 2, 3, …18 are put in a box and mixed thoroughly. A card is drawn at random from the box. Find the probabilities that the card bears
(i) an even number
(ii) a number divisible by 2 or 3
Ans. Total no. of possible outcomes = 18
(i) Favorable cases are 2,4,6,8,10,12,14,16,18 i.e., 9 in number
Required probability = (ii) Favorable cares are 2,3,4,6,8,9,10,12,14,15,16,18 i.e., 12 in number
Required probability = 6. A bag contains 5 red balls, 4 green balls and 7 white balls. A ball is drawn at random from the box. Find the probability that the ball drown is
(a) white
(b) neither red nor white
Ans . Total number of balls in the bag = 5+4+7=16 Total number of possible outcomes = 16
(a) Favourable outcomes for a white ball = 7
Required probability = (b) Favourable outcomes for neither red nor white ball=Number of green balls =4 Required probability = 7. A box contains 20 balls bearing numbers 1,2,3,4,…20. A ball is drawn at random from the box, what is the probability that the number on the ball is
(i) an odd number
(ii) divisible by 2 or 3
(iii) prime number
Ans. Total number of outcomes = 20
(i) Favorable outcomes are 1,3,5,7,9,11,13,15,17,19 i.e., 10 in number. Required probability = (ii) Number “divisible by 2” are 2,4,6,8,10,12,14,16,18,20 i.e., 10 in number
Numbers “divisible by 3 are 3,6,9,12,15,18. i.e., 6 in number
Numbers “divisible by 2 or 3 are 6,12,18 i.e., 3 in number. Numbers divisible by “2 or 3”=10+6 – 3=13
Favourable outcomes = 13 Required probability = (iii) Prime numbers are 2,3,5,7,11,13,17,19 i.e., 8 in number
Favourable outcomes = 8
Required probability = 8. A bag contains 5 red and some blue balls,
(i) if probability of drawing a blue ball from the bag is twice that of a red ball,find the number of blue balls in the bag.
(ii) if probability of drawing a blue ball from the bag is four times that of a red ball,find the number of blue balls in the bag.
Ans. Let number of blue balls = Total number of balls = Probability of red ball = Probability of blue ball = By given condition,
(i) No. of blue balls = 10
(ii) Here,  Hence, number of blue balls = 20

9. A box contains 3 blue marbles, 2 white marbles. If a marble is taken out at random from the box, what is the probability that it will be a white one? Blue one? Red one?
Ans. Total no. of possible outcomes = 3+2+4 = 9
No. of favourable outcomes for white marbles = 2
Required probability = No. of favourable outcomes for blue marbles = 3
Required probability = No. of favourable outcomes for red marbles =4
Required probability = 10. The integers from 1 to 30 inclusive are written on cards ( one number on one card). These card one put in a box and well mixed. Joseph picked up one card. What is the probability that his card has
(i) number 7
(ii) an even number
(iii) a prime number
Ans. Total no. of possible outcomes = 30
(i) P (the no.7) = (ii) Even no. are 2,4,6,8,10,12,14,16,18,20,22,24,26,28,30
Favourable outcomes = 15
Required probability = (iii) Prime numbers from 1 to 30 are 2,3,5,7,11,13,17,19,23,29}
No. of favourable outcomes = 10
Required probability = 11. A bag contains lemon flavored candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
(i) an orange flavored candy?
(ii) alemon flavored candy?
Ans. The bag has lemon flavored candies only.
(i) P(an orange flavored candy) = (ii) P (a lemon flavored candy ) = 12. A bag contains 6 red balls and some blue balls. If the probability of drawing a blue ball from the bag is twice that of a red, find the number of blue balls in the bag.
Ans. Suppose no. of blue bolls = Total no. of balls = Probability of blue balls = Probability of red balls = According to, question, Hence, no. of blue balls = 12

13. A bag contains 5 red, 4 black and 3 green balls. A ball is taken out of the bag at random, find the probability that the selected ball is
(i) of red colour,
(ii) not of green colour.
Ans. Total number of balls in the bag = 5+4+3 Total number of outcomes = 12
(i) No. of red balls = 5 Required probability for a red ball = (ii) favourable cases for non green ball = 12 – 3=9 Required probability for a non green ball = 14. From a well shuffled pack of 52 cards, black aces and black queens are removed. From the remaining cards a card is drawn at random, find the probability of drawing a king or a queen.
Ans. Total number of cards = 52
Number of black aces = 2
Number of black queens = 2
Cards left = 52 – 2 – 2=48 Total number of equally likely cases = 48
Number of kings and queens left in the 48 cards = 4+2=6
Favourable cases = 6 Required probability = = 15. Which of the following experiments have equally likely outcomes? Explain.
(i) A driver attempts to start a car. The car starts or does not start.
(ii) A player attempts to shoot a basket ball, she/he shoots or misses the shot.
(iii) A baby is born. It is a boy or a girl.
Ans. (i) A driver attempts to start a car,thencarstarts or does not start are not equally likely.
(ii) A player attempts to shoot a basket ball, she/he shoots or misses the shoot are not equally likely.
(iii) A baby is born,it is a boy or a girl is an equally likely event.

16. Find the probability that a number selected at random from the numbers 1,2,3,…35 is a
(i) prime number,
(ii) multiple of 7,
(iii) multiple of 3 or 5.
Ans. (i) Prime number are 2,3,5,7,11,13,17,19,23,29,31, which are 11 in number Total number of outcomes = 35
P (Prime number) = (ii) Multiples of ‘7’ are 7,14,21,28,35, which are 5 in number
P (a multiple of 7) = (iii) Multiple of ‘3’ are 3,6,9,…33, which are 11 in numbers
Multiple of 5 are 5,10,15,…35, which are 7 in number
Multiple of ‘3’ and ‘5’ are 15, 30, which are 2 in number Multiple of 3 or 5 = 11+7 – 2=16 P (Multiple of 3 or 5) = 