CBSE Class 10 Maths Chapter-15 Probability – Free PDF Download
Free PDF download of Important Questions with Answers for CBSE Class 10 Maths Chapter 15 – Probability prepared by expert Maths teachers from latest edition of CBSE(NCERT) books only by CoolGyan to score more marks in CBSE board examination.
CBSE Class 10 Maths Chapter-15 Probability Important Questions
CBSE Class 10 Maths Important Questions Chapter 15 – Probability
2 Mark Questions
1. 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 basketball. She/he shoots or misses the shot.
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
(iv) A baby is born. It is a boy or a girl.
Ans. (i) In the experiment, “A driver attempts to start a car. The car starts or does not start”, we are not justified to assume that each outcome is as likely to occur as the other. Thus, the experiment has no equally likely outcomes.
(ii) In the experiment, “A player attempts to shoot a basket ball. She/he shoots or misses the shot”, we are not justified to assume that each outcome is as likely to occur as the other. Thus, the experiment has no equally likely outcomes.
(iii) In the experiment “A trial is made to answer a true-false question. The answer is right or wrong.” We know, in advance, that the result can lead in one of the two possible ways – either right or wrong. We can reasonably assume that each outcome, right or wrong, is likely to occur as the other.
Thus, the outcomes right or wrong are equally likely.
(iv) In the experiment, “A baby is born, It is a boy or a girl”. We know, in advance that the outcome can lead in one of the two possible outcomes – either a boy or a girl. We are justified to assume that each outcome, boy or girl, is likely to occur as the other. Thus, the outcomes boy or girl are equally likely.
2. Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?
Ans. The tossing of a coin is considered to be a fair way of deciding which team should get the ball at the beginning of a football game as we know that the tossing of the coin only land in one of two possible ways – either head up or tail up. It can reasonably be assumed that each outcome, head or tail, is as likely to occur as the other, i.e., the outcomes head and tail are equally likely. So the result of the tossing of a coin is completely unpredictable.
3. A bag contains lemon flavoured candles only. Malini takes out one candy without looking into the bag. What is the probability that she takes out:
(i) an orange flavoured candy?
(ii) a lemon flavoured candy?
Ans. (i) Consider the event related to the experiment of taking out of an orange flavoured candy from a bag containing only lemon flavoured candies.
Since no outcome gives an orange flavoured candy, therefore, it is an impossible event so its probability is 0.
(ii) Consider the event of taking a lemon flavoured candy out of a bag containing only lemon flavoured candies. This event is a certain event so its probability is 1.
4. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is:
(i) red?
(ii) not red?
Ans. There are 3 + 5 = 8 balls in a bag. Out of these 8 balls, one can be chosen in 8 ways.
Total number of elementary events = 8
(i) Since the bag contains 3 red balls, therefore, one red ball can be drawn in 3 ways.
Favourable number of elementary events = 3
Hence P (getting a red ball) = 
(ii) Since the bag contains 5 black balls along with 3 red balls, therefore one black (not red) ball can be drawn in 5 ways.
Favourable number of elementary events = 5
Hence P (getting a black ball) = 
5. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be:
(i) red?
(ii) white?
(iii) not green?
Ans. Total number of marbles in the box = 5 + 8 + 4 = 17
Total number of elementary events = 17
(i) There are 5 red marbles in the box.
Favourable number of elementary events = 5
P (getting a red marble) = 
(ii) There are 8 white marbles in the box.
Favourable number of elementary events = 8
P (getting a white marble) = 
(iii) There are 5 + 8 = 13 marbles in the box, which are not green.
Favourable number of elementary events = 13
P (not getting a green marble) = 
6. A piggy bank contains hundred 50 p coins, fifty Re. 1 coins, twenty Rs. 2coins and ten Rs. 5 coins. If it is equally likely that of the coins will fall out when the bank is turned upside down, what is the probability that the coin:
(i) will be a 50 p coin?
(ii) will not be a Rs.5 coin?
Ans. Total number of coins in a piggy bank = 100 + 50 + 20 + 10 = 180
Total number of elementary events = 180
(i) There are one hundred 50 coins in the piggy bank.
Favourable number of elementary events = 100
P (falling out of a 50 p coin) = 
= 
(ii) There are 100 + 50 + 20 = 170 coins other than Rs. 5 coin.
Favourable number of elementary events = 170
P (falling out of a coin other than Rs. 5 coin) = 
= 
7. Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fishes and 8 female fishes (see figure). What is the probability that the fish taken out is a male fish?
Ans. Total number of fish in the tank = 5 + 8 = 13
Total number of elementary events = 13
There are 5 male fishes in the tank.
Favourable number of elementary events = 5
Hence, P (taking out a male fish) = 
8. Five cards – then ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen?
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?
Ans. Total number of favourable outcomes = 5
(i) There is only one queen.
Favourable outcome = 1
Hence, P (the queen) = 
(ii) In this situation, total number of favourable outcomes = 4
(a) Favourable outcome = 1
Hence, P (an ace) = 
(b) There is no card as queen.
Favourable outcome = 0
Hence, P (the queen) = 
9. (i) A lot of 20 bulbs contains 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
Ans. (i) Total number of favourable outcomes = 20
Number of favourable outcomes = 4
Hence P (getting a defective bulb) = 
(ii) Now total number of favourable outcomes = 20 – 1 = 19
Number of favouroable outcomes = 19 – 4 = 15
Hence P (getting a non-defective bulb) = 
10. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears (i) a two-digit number
(ii) a perfect square number
(iii) a number divisible by 5.
Ans. Total number of favourable outcomes = 90
(i) Number of two-digit numbers from 1 to 90 are 90 – 9 = 81
Favourable outcomes = 81
Hence, P (getting a disc bearing a two-digit number) = 
(ii) From 1 to 90, the perfect squares are 1, 4, 9, 16, 25, 36, 49, 64 and 81.
Favourable outcomes = 9
Hence P (getting a perfect square) = 
(iii) The numbers divisible by 5 from 1 to 90 are 18.
Favourable outcomes = 18
Hence P (getting a number divisible by 5) = 
11. A child has a die whose six faces show the letters as given below:
A, B, C, D, E, A
The die is thrown once. What is the probability of getting:
(i) A?
(ii) D?
Ans. Total number of favourable outcomes = 6
(i) Number of favourable outcomes = 2
Hence P (getting a letter A) = 
(ii) Number of favourable outcomes = 1
Hence P (getting a letter D) = 
12. Suppose you drop a die at random on the rectangular region shown in the figure given on the next page. What is the probability that it will land inside the circle with diameter 1 m?
Ans. Total area of the given figure (rectangle) = 3 x 2 = 6 m2
And Area of circle = 
= 
= 
m2
Hence, P (die to land inside the circle) = 
13. A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that:
(i) she will buy it?
(ii) she will not buy it?
Ans. Total number of favourable outcomes = 144
(i) Number of non-defective pens = 144 – 20 = 124
Number of favourable outcomes = 124
Hence P (she will buy) = P (a non-defective pen) = 
(ii) Number of favourable outcomes = 20
Hence P (she will not buy) = P (a defective pen) = 
14. 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 there be 
blue balls in the bag.
Total number of balls in the bag = 
Now, P1 = Probability of drawing a blue ball = 
And P2 = Probability of drawing a blue ball = 
But according to question, P1 = 2P2
= 2 x
=2
Hence, there are 10 blue balls in the bag.
15. 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 
Ans. There are 12 balls in the box.
Therefore, total number of favourbale outcomes = 12
The number of favourable outcomes =
Therefore, P1 = 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, P2 = 2P1
= 2 x
=2
16. 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.
Ans. Here, Total number of favourable outcomes = 24
Let there be green marbles.
Therefore, Favourable number of outcomes =
P(G) = 
But P(G) = 
= 16
Therefore, number of green marbles are 16
And number of blue marbles = 24 – 16 = 8
17. Why is tossing a coin considered is the way of deciding which team should get the ball at the beginning of a football match?
Ans. 
Probability of head
Probability of tail 
i.e.
Probability of getting head and tail both are same.
Tossing a coin considered tobe fair way.
18. An unbiased die is thrown, what is the probability of getting an even number?
Ans. Total number of outcomes are 1,2,3,4,5 and 6, which are 6 in number favourable case = 1 
Required probability = 
19. Two unbiased coins are tossed simultaneously, find the probability of getting two heads.
Ans. Total number of outcomes are HH, HT, TH, TT, which are 4 in numbers
Favourable outcomes = HH, which is only
Required probability = 
20. One card is drawn from a well shuffled deck of 52 cards. Find the probability of getting a jack of hearts.
Ans. Total number of outcomes = 52
Favorable cases = 1 [There is only one jack of hearts]
Required probability = 
21. 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 fails and loses otherwise. Calculate the probability that Hanif will lose the game.
Ans. Since a coin is tossed 3 times
Possible outcomes are = {HHH,HHT,HTH,THH,HTT,THT,TTH,TTT}
3 heads and 3 tails {HHH,TTT}
(3 heads and 3 tails) = 2
P(Hanif will win the game) = 
P (Hanif will lose the game) = 
22. Gopy buys a fish from a shop for his aquarium. The shopkeeper take out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish?
Ans. Total no. of fishes = 5+8=13
No. of male fishes = 5
P(male fish) = 
23. A lot consists of 144 ball pens of which 20 are defective and the others are food. Arti will buy a pen if it is good but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that
(i) she will buy it
(ii) she will not buy it?
Ans. Total no. of ball pens = 144
Number of defective pens = 20
Number of good pens = 144 – 20=124
(i) P(she will buy it) = 
(ii) P (she will not buy) = 
24. Harpreet tosses two different coins simultaneously (say one is of Rs 1 and other is Rs 2), what is the probability that she gets “at least one head”?
Ans. For tossing 2 coins
Total possible outcomes are {HH,TT,TH,HT}= 4
Favourable outcomes = at least one head= {TH,HT,HH}=3
Required probability = 
25. Why is tossing a coin considered is the way of deciding which team should get the ball at the beginning of a football match?
Ans. Probability of head
Probability of tail
i.e.
Probability of getting head and tail both are same.
Tossing a coin considered to be fair way.
26. Two unbiased coins are tossed simultaneously, find the probability of getting two heads.
Ans. Total number of outcomes are HH, HT, TH, TT, which are 4 in numbers
Favourable outcomes = HH, which is only
Required probability =
27. One card is drawn from a well shuffled deck of 52 cards. Find the probability of getting a jack of hearts.
Ans. Total number of outcomes = 52
Favorable cases = 1 [There is only one jack of hearts]
Required probability =
28. If two dice are thrown once, find the probability of getting 9.
Ans. Total number of possible outcomes of throwing two dice = 
Number of outcomes of getting 9 i.e., (3+6),(4+5),(5+4),(6+3) = 4
Required probability 
29. A card is drawn from a well shuffled deck of playing cards. Find the probability of getting a face card.
Ans. Total number of possible outcomes = 52
Favourable outcomes = 4+4+4=12[4 jack, 4 queen, 4 king]
Required probability = 
30. What is the probability of having 53 Mondays in a leap year?
Ans. Total number of days in a leap year = 366
This contains 52 weeks and 2 days
The remaining two days may beMT, TW,WTh, ThF, FS, SS, SM
Favourable cases are MT, SM i.e., 2 out of 7 cases
Required probability = 
31. Cards bearing numbers 3 to 20 are placed in a bag and mixed thoroughly. A card is taken out from the bag at random, what is the probability that the number on the card taken out is an even number?
Ans. Total number of outcomes = 20 – 3=17
Cards in the box having even numbers are 4,6,8,10,12,14,16,18,20, which are 9 in number
favourable outcomes= 9
P (an even number) = 