NCERT Solutions for Class 10 Maths Exercise 15.1 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. Complete the statements:
(i) Probability of event E + Probability of event “not E” = _______________
(ii) The probability of an event that cannot happen is _______________. Such an event is called _______________.
(iii) The probability of an event that is certain to happen is _______________. Such an event is called _______________.
(iv) The sum of the probabilities of all the elementary events of an experiment is _______________.
(v) The probability of an event is greater than or equal to _______________ and less than or equal to _______________.
(ii) 0, impossible event
(iii) 1, sure or certain event
(iv) 1
(v) 0, 1
2. 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.
(ii) In the experiment, “A player attempts to shoot a basketball. She/he shoots or misses the shot”, the outcome depends upon many factors e.g. quality of player. 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 to 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 there are only 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.
3. 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?
4. Which of the following cannot be the probability of an event:
(A) 
(B) 
(C) 15%
(D) 0.7
0 
P(E) 1
cannot be the probability of an event.
5. If P(E) = 0.05, what is the probability of ‘not E’?
P (not E) = 1 – P(E) = 1 – 0.05 = 0.95
6. 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?
(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.
7. It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?
P(E) = 0.992
But P(E) + P
= 1
P = 1 – P(E) = 1 – 0.992 = 0.008
8. 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?
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) = 
9. 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?
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) = 
10. 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?
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) = 
= 
11. 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?
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) = 
12. 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?
Total number of possible 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
13. 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.
(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) =
14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting:
(i) a king of red colour
(ii) a face card
(iii) a red face card
(iv) the jack of hearts
(v) a spade
(vi) the queen of diamonds.
(i) There are two suits of red cards, i.e., diamond and heart. Each suit contains one king.
Favourable outcomes = 1
Hence, P (a king of red colour) = 
(ii) There are 12 face cards in a pack.
Favourable outcomes = 12
Hence, P (a face card) = 
(iii) There are two suits of red cards, i.e., diamond and heart. Each suit contains 3 face cards.
Favourable outcomes = 
= 6
Hence, P (a red face card) = 
(iv) There are only one jack of heart.
Favourable outcome = 1
Hence, P (the jack of hearts) = 
(v) There are 13 cards of spade.
Favourable outcomes = 13
Hence, P (a spade) = 
(vi) There is only one queen of diamonds.
Favourable outcome = 1
Hence, P (the queen of diamonds) =
15. 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?
(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) = 
16. 12 defective pens are accidently mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Number of favourable outcomes = 132
Hence, P (getting a good pen) = 
17. (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?
Number of favourable outcomes = 4
Hence P (getting a defective bulb) = 
(ii) Now total number of possible outcomes = 20 – 1 = 19
Number of favouroable outcomes = 19 – 4 = 15
Hence P (getting a non-defective bulb) = 
18. 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.
(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) = 
19. 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?
(i) Number of favourable outcomes = 2
Hence P (getting a letter A) = 
(ii) Number of favourable outcomes = 1
Hence P (getting a letter D) = 
20. 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?
And Area of circle = 
= 
= 
m2
Hence, P (die to land inside the circle) = 
21. 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?
(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) = 
22. Refer to example 13.
(i) Complete the following table:
(ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore each of them has a probability 
Do you agree with this argument? Justify your answer.
(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 number of favourable outcomes = 36
(i) Favourable outcomes of getting the sum as 3 = 2
Hence P (getting the sum as 3) = 
Favourable outcomes of getting the sum as 4 = 3
Hence P (getting the sum as 4) = 
Favourable outcomes of getting the sum as 5 = 4
Hence P (getting the sum as 5) = 
Favourable outcomes of getting the sum as 6 = 5
Hence P (getting the sum as 6) = 
Favourable outcomes of getting the sum as 7 = 6
Hence P (getting the sum as 7) = 
Favourable outcomes of getting the sum as 9 = 4
Hence P (getting the sum as 9) =
Favourable outcomes of getting the sum as 10 = 3
Hence P (getting the sum as 10) =
Favourable outcomes of getting the sum as 11 = 2
Hence P (getting the sum as 11) =
(ii) I do not agree with the argument given here. Justification has already been given in part (i).
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.
HHH, HHT, HTH, THH, TTH, HTT, THT, TTT
Therefore, Total number of possible outcomes = 8
Number of favourable outcomes = 6
Hence required probability =
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?
(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)
Therefore, Total number of possible outcomes = 36
Now consider the following events:
A = first throe shows 5 and B = second throw shows 5
Therefore, the number of favourable outcomes = 6 in each case.
P(A) = 
and P(B) =
P
and P
Required probability = 
(ii) Let S be the sample space associated with the random experiment of throwing a die twice. Then, n(S) = 36
A
B = first and second throw shoe 5, i.e. getting 5 in each throw.
We have, A = (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
And B = (1, 5) (2, 5) (3, 5) (4, 5) (5, 5) (6, 5)
P(A) = , P(B) = and P(AB) = 
Required probability = Probability that at least one of the two throws shows 5
= P (A
B) = P(A) + P(B) – P(AB)
= 
25. Which of the following arguments are correct and which are not correct? Give reasons for your answer:
(i) 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 
(ii) 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 
(ii) Correct: The two outcomes considered in the question are equally likely.