NCERT Solutions Maths Class 10 Exercise 15.1

# NCERT Solutions Maths Class 10 Exercise 15.1

## Maths Class 10 Exercise 15.1

1. Complete the following statements:

(i) Probability of an event E + Probability of event ‘not E’ = 1.

(ii) The probability of an event that cannot happen is 0. Such an event is called impossible event.

(iii) The probability of an event that is certain to happen is 1. Such an event is called sure or certain event.

(iv) The sum of the probabilities of all the elementary events of an experiment is 1.

(v) The probability of an event is greater than or equal to 0 and less than or equal to        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.

(iv) A baby is born. It is a boy or a girl.

Solution: (i) In the experiment, “A driver attempts to start a car. The car starts or does not start”, it is not sure 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 basketball. She/he shoots or misses the shot”, it is not sure 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”, it is clear that the result can lead in one of the two possible answers – either right or wrong. We can clearly justify that each outcome, right or wrong, is likely to occur as the other. Thus, the experiment has equally likely outcomes.

(iv) In the experiment, “A baby is born. It is a boy or a girl”, it is clear that the outcome can lead in one of the two possible outcomes – either a boy or a girl. We can clearly justify that each outcome, a boy or a girl, is likely to occur as the other. Thus, the experiment has equally likely outcomes.

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?

Solution: 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 because it is clear that the tossing of the coin only land in one of two possible sides – either head up or tail up. It can be justified 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 a coin is completely unpredictable.

4. Which of the following cannot be the probability of an event?

(A) 2/3                  (B) ­–1.5               (C) 15%                    (D) 0.7

Solution: (B) Since the probability of an event E is a number P(E) such that

0 ≤ P(E) ≤ 1

It means that the probability of an event is greater or equal to 0 but less than or equal to 1. Here, –1.5 is less than 0 because of negative sign.

Therefore, –1.5 cannot be the probability of an event.

5. If P(E) = 0.05, what is the probability of ‘not E’?

Solution: We know that, P(E) + P(not E) = 1

or P(not E) = 1 – P(E)

= 1 – 0.05

= 0.95

Therefore, the probability of ‘not E’ is 0.95.

6. A bag contains lemon flavoured candies 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?

Solution: (i) Consider the event E 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. Hence, P(E) = 0.

(ii) Consider the event E of taking a lemon flavoured candy out from a bag containing only lemon flavoured candies. This event is a sure or a certain event. So, its probability is 1. Hence, P(E) = 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?

Solution: Let E be the event of having the same birthday, then ‘not E’ is the event of not having the same birthday.

It is given that, P(not E) = 0.992

But P(E) + P(not E) = 1

P(E) = 1 – P(not E)

= 1 – 0.992

= 0.008

Hence, the probability that the 2 students have the same birthday is 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?

Solution: There are a total of 3 + 5 = 8 balls in a bag. Out of these 8 balls, one can be drawn in 8 ways.

Total number of possible outcomes = 8

(i) Since the bag contains 3 red balls, therefore, one red ball can be drawn in 3 ways.

Number of favourable outcomes = 3

Hence, P(getting a red ball) = 3/8

(ii) Since the bag contains 5 black balls along with 3 red balls, therefore, 5 balls are not red balls. One black (not red) ball can be drawn in 5 ways.

Number of favourable outcomes = 5

Hence, P(getting not a red ball) = 5/8

Note: We can simply solve this question as follows:

If P(E) denotes the probability of getting a red ball, then the probability of getting not a red ball, i.e., P(not E) can be calculated by the formula P(not E) = 1 P(E).

Thus, P(getting not a red ball) = 1 P(getting a red ball)

P(getting not a red ball) = 1 3/8 = 5/8

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?

Solution: Total number of marbles in the box = 5 + 8 + 4 = 17

Total number of possible outcomes = 17

(i) There are 5 red marbles in the box.

Number of favourable outcomes = 5

P(getting a red marble) = 5/17

(ii) There are 8 white marbles in the box.

Number of favourable outcomes = 8

P(getting a white marble) = 8/17

(iii) There are 5 + 8 = 13 marbles in the box, which are not green.

Number of favourable outcomes = 13

P(getting not a green marble) = 13/17

10. A piggy bank contains hundred 50 p coins, fifty Re. 1 coins, twenty Rs. 2 coins and ten Rs. 5 coins. If it is equally likely that one 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?

Solution: Total number of coins in the piggy bank = 100 + 50 + 20 + 10 = 180

Total number of possible outcomes = 180

(i) There are one hundred 50 p coins in the piggy bank.

Number of favourable outcomes = 100

P(falling out a 50 p coin) =  100/180 =  5/9

(ii) There are 100 + 50 + 20 = 170 coins, which are not Rs. 5 coins.

Number of favourable outcomes = 170

So, P(falling out a coin which is not a Rs. 5 coin) = 170/180 = 17/18

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 fish and 8 female fish (see figure). What is the probability that the fish taken out is a male fish?

Solution: Total number of fish in the tank = 5 + 8 = 13

Total number of possible outcomes = 13

There are 5 male fish in the tank.

Number of favourable outcomes = 5

Hence, P(taking out a male fish) = 5/13

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?

Solution: Out of 8 numbers, an arrow can point any of the numbers in 8 ways.

Total number of possible outcomes = 8

(i) Since there are only one 8.

Number of favourable outcomes = 1

Hence, P(arrow points at 8) = 1/8

(ii) Since there are 4 odd numbers, i.e., 1, 3, 5, 7.

Number of favourable outcomes = 4

Hence, P(arrow points at an odd number) = 4/8 = 1/2

(iii) Since the numbers greater than 2 are 3, 4, 5, 6, 7, 8.

Number of favourable outcomes = 6

Hence, P(arrow points at a number > 2) = 6/8 = 3/4

(iv) Since all the 8 numbers are less than 9.

Number of favourable outcomes = 8

Hence, P(arrow points at a number < 9) = 8/8 = 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.

Solution: Total number of possible outcomes when throwing a dice = 6

(i) On a dice, the prime numbers are 2, 3 and 5.

Therefore, the number of favourable outcomes = 3

Hence, P(getting a prime number) = 3/6 = 1/2

(ii) On a dice, the number lying between 2 and 6 are 3, 4, 5.

Therefore, the number of favourable outcomes = 3

Hence, P(getting a number lying between 2 and 6) = 3/6 =1/2

(iii) On a dice, the odd numbers are 1, 3 and 5.

Therefore, the number of favourable outcomes = 3

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

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

(vi) the queen of diamonds.

Solution: Total number of possible outcomes = 52

(i) There are two suits of red cards, i.e., diamond and heart. Each suit contains one king. So, there are 2 kings of red colour.

Number of favourable outcomes = 2

Hence, P(a king of red colour) = 2/52 = 1/26

(ii) Since there are 3 face cards in each suit, therefore, a total of 12 face cards are there in all four suits.

Number of favourable outcomes = 12

Hence, P(a face card) = 12/52 = 3/13

(iii) There are two suits of red cards, i.e., diamond and heart. Each suit contains 3 face cards.

Number of favourable outcomes = 2 × 3 = 6

Hence, P(a red face card) = 6/52 = 3/26

(iv) There is only one jack of heart.

Number of favourable outcomes = 1

Hence, P(the jack of hearts) = 1/52

(v) There are 13 cards of spade.

Number of favourable outcomes = 13

Hence, P(a spade) = 13/52 = 1/4

(vi) There is only one queen of diamonds.

Number of favourable outcomes = 1

Hence, P(the queen of diamonds) = 1/52

15. Five cards – the 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?

Solution: Total number of possible outcomes = 5

(i) There is only one queen.

Number of favourable outcomes = 1

Hence, P(the queen) = 1/5

(ii) If the queen is put aside, total number of possible outcomes = 4

(a) Number of favourable outcomes = 1

Hence, P(an ace) = 1/4

(b) Since the queen is put aside, there is no card as queen.

Number of favourable outcomes = 0

Hence, P(the queen) = 0/4 = 0

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.

Solution: Total number of possible outcomes = 132 + 12 = 144

Number of favourable outcomes = 132

Hence, P(getting a good pen) = 132/144 = 11/12

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?

Solution: (i) Total number of possible outcomes = 20

Number of favourable outcomes = 4

Hence, P(getting a defective bulb) = 4/20 = 1/5

(ii) If the bulb drawn in (i) is not defective, then in the lot, number of defective bulb is 4 and number of not defective bulb is 15.

Now, total number of possible outcomes = 20 – 1 = 19

Number of favourable outcomes = 19 – 4 = 15

Hence, P(getting a non-defective bulb) = 15/19

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.

Solution: Total number of possible outcomes = 90

(i) Number of two-digit numbers from 1 to 90 are 90 – 9 = 81

Number of favourable outcomes = 81

Hence, P(getting a disc bearing a two-digit number) = 81/90 = 9/10

(ii) From 1 to 90, the perfect square numbers are 1, 4, 9, 16, 25, 36, 49, 64 and 81.

Number of favourable outcomes = 9

Hence, P(getting a perfect square number) = 9/90 = 1/10

(iii) From 1 to 90, the numbers divisible by 5 are 5, 10, 15, 20, ……., 90

There are total 18 numbers which are divisible by 5.

Number of favourable outcomes = 18

Hence, P(getting a number divisible by 5) = 18/90 = 1/5

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?

Solution: Total number of possible outcomes = 6

(i) Since the letter A appears on the two faces.

Number of favourable outcomes = 2

Hence, P(getting a letter A) = 2/6 = 1/3

(ii) Since the letter D appears on only one face.

Number of favourable outcomes = 1

Hence, P(getting a letter D) = 1/6

20. Suppose you drop a die at random on the rectangular region shown in the following figure. What is the probability that it will land inside the circle with diameter 1 m?

Solution: Area of the rectangle = 3 × 2 = 6 m2

And area of the circle = πr2 = π(1/2)2 = π/4 m2

Total possible area to land the die = 6 m2

Total favourable area to land the die = π/4 m2

Hence, P(die to land inside the circle) = π/4/6 = π/24

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

(ii) she will not buy it?

Solution: Total number of possible 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) = 124/144 = 31/36

(ii) Number of favourable outcomes = 20

Hence, P(she will not buy) = P(a defective pen) = 20/144 =  5/36

22. Refer to example 13.

(i) Complete the following table:

 Event: ‘Sum on 2 dice’ 2 3 4 5 6 7 8 9 10 11 12 Probability 1/36 5/36 1/36

(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 1/11. Do you agree with this argument? Justify your answer.

Solution: Total possible outcomes of throwing two dice are:

(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 possible outcomes = 36

(i) Favourable outcomes of getting the sum as 3, i.e., (1, 2) and (2, 1) = 2

Hence, P(getting the sum as 3) = 2/36 = 1/18

Favourable outcomes of getting the sum as 4, i.e., (1, 3), (3, 1) and (2, 2) = 3

Hence, P(getting the sum as 4) = 3/36 = 1/12

Favourable outcomes of getting the sum as 5 = 4

Hence, P(getting the sum as 5) = 4/36 = 1/9

Favourable outcomes of getting the sum as 6 = 5

Hence, P(getting the sum as 6) = 5/36

Favourable outcomes of getting the sum as 7 = 6

Hence, P(getting the sum as 7) = 6/36 = 1/6

Favourable outcomes of getting the sum as 9 = 4

Hence, P(getting the sum as 9) = 4/36 = 1/9

Favourable outcomes of getting the sum as 10 = 3

Hence, P(getting the sum as 10) = 3/36 = 1/12

Favourable outcomes of getting the sum as 11 = 2

Hence, P(getting the sum as 11) = 2/36 = 1/18

 Event: ‘Sum on 2 dice’ 2 3 4 5 6 7 8 9 10 11 12 Probability 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36

(ii) I do not agree with the argument given by the student. 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.

Solution: The outcomes when a coin is tossed 3 times:

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

Total number of possible outcomes = 8

Number of favourable outcomes = 6

Hence, P(Hanif will lose the game) = 6/8 = 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?

Solution: (i) The outcomes when a dice is thrown is twice:

(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 possible outcomes = 36

Now, consider the event E in which 5 will come up either time.

Then, E = (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (1, 5) (2, 5) (3, 5) (4, 5) (6, 5)

Total number of outcomes favourable to E = 11

P(5 will come up either time) = P(E) = 11/36

P(5 will not come up either time) = P(not E) = 1 – 11/36 = 25/36

(ii) Total number of possible outcomes = 36

Now, consider the event E in which 5 will come up at least once.

Then, E = (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (1, 5) (2, 5) (3, 5) (4, 5) (6, 5)

Total number of outcomes favourable to E = 11

P(5 will come up at least once) = P(E) = 11/36

###### 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 1/3.

(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 1/2.

Solution: (i) Incorrect: We can classify the outcomes like this but they are not then, ‘equally likely’. Reason is that ‘one of each’ can result in two ways – from a head on first coin and tail on the second coin or from a tail on the first coin and head on the second coin. This makes it twice as likely as two heads (or two tails).

(ii) Correct: The two outcomes considered in the question are equally likely.