﻿ Class 10 NCERT Math Solution
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Exercise - 15.1

Question-1 :-  Complete the following statements:
(i) Probability of an event E + Probability of the 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 ........

Solution :-
```(i) Probability of an event E + Probability of the event ‘not E’ = 1
(ii) The probability of an event that cannot happen is 0 Such an event is called imposible event.
(iii) The probability of an event that is certain to happen is 1 Such an event is called sure 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.
```

Question-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) It is not an equally likely event, as it depends on various factors such as whether
the car will start or not. And factors for both the conditions are not the same.

(ii) It is not an equally likely event, as it depends on the player’s ability and there is

(iii) It is an equally likely event.

(iv) It is an equally likely event.
```

Question-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 :-
```  When we toss a coin, the possible outcomes are only two, head or tail, which are
equally likely outcomes. Therefore, the result of an individual toss is completely
unpredictable.
```

Question-4 :-  Which of the following cannot be the probability of an event?
(A) 2/3  (B) –1.5  (C) 15%  (D) 0.7

Solution :-
```  Probability of an event (E) is always greater than or equal to 0. Also, it is always less
than or equal to one. This implies that the probability of an event cannot be negative
or greater than 1. Therefore, out of these alternatives, −1.5 cannot be a probability
of an event.
```

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

Solution :-
```  Given :
P(E) = 0.05
P(E not) = 1 - P(E)
P(E not) = 1 - 0.05
P(E not) = 0.95
Therefore, the probability of ‘not E’ is 0.95.
```

Question-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) The bag contains lemon flavoured candies only. It does not contain any orange
flavoured candies. This implies that every time, she will take out only lemon
flavoured candies. Therefore, event that Malini will take out an orange flavoured
candy is an impossible event.
Hence, P (an orange flavoured candy) = 0
```
```(ii) As the bag has lemon flavoured candies, Malini will take out only lemon flavoured
candies. Therefore, event that Malini will take out a lemon flavoured candy is a sure
event.
P(a lemon flavoured candy) = 1
```

Question-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 :-
```  Probability that two students are not having same birthday P(E not ) = 0.992
Probability that two students are having same birthday
P(E) = 1 − P(E not)
= 1 − 0.992
= 0.008
```

Question-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 :-
```  Let R and B are be the event that red ball and black ball respectively.
No. of red balls = 3
No. of black balls = 5
Total number of balls in the bag = 3 + 5 = 8

(i) red
P(R) = 3/8

(ii) not red
P(R not) = 1 - P(R)
= 1 - 3/8
= 5/8
```

Question-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 :-
```  Let R, W and G are be the event of red, white and green marbles respectively.
No. of red marbles = 5
No. of white marbles = 8
No. of green marbles = 4
Total number of marbles = 5 + 8 + 4 = 17

(i) red
P(R) = 5/17

(ii) white
P(W) = 8/17

(iii) not green
P(G) = 4/17
P(G not) = 1 - 4/17
P(G not) = 13/17
```

Question-10 :-  A piggy bank contains hundred 50p coins, fifty ₹ 1 coins, twenty ₹ 2 coins and ten ₹ 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 ₹ 5 coin ?

Solution :-
```  Let E, F, G and H are be the event of 50p, ₹ 1, ₹ 2 and ₹ 5 coins respectively.
No.of coins of 50p = 100
No. of coins of ₹ 1 = 50
No. of coins of ₹ 2 = 20
No. of coins of ₹ 5 = 10
Total number of coins in a piggy bank = 100 + 50 + 20 + 10 = 180

(i) 50 p coin
P(E) = 100/180
= 5/9

(ii) not ₹ 5 coin
P(H) = 10/180
= 1/18
P(H not) = 1 - 1/18
= 17/18
```

Question-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 Fig.). What is the probability that the fish taken out is a male fish? Solution :-
```  Let M and F are be the event of male and female fish respectively.
No. of male fishes = 5
No. of female fishes = 8
Total number of fishes in a tank = Number of male fishes + Number of female fishes = 5 + 8 = 13.

Probability of male fishes = P(M)
P(M) = 5/13
```

Question-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 Fig. 15.5 ), 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 :-
```  Total number of possible outcomes = 8

(i) Probabiity of getting 8 = P(8)
No. of outcomes 8 = 1
P(8) = 1/8

(ii) Probability of gettingan odd no. = P(O)
No. of outcomes of odd no. (i.e. 1, 3, 5, 7) = 4
P(O) = 4/8 = 1/2

(iii) Probability of getting a number greater than 2 = P(G)
No. of outcomes of a number greater than 2 (i.e. 3, 4, 5, 6, 7, 8) = 6
P(G) = 6/8 = 3/4

(iv) Probability of a number less than 9 = P(L)
No. of outcomes of a number less than 9 (i.e. 1, 2, 3, 4, 5, 6, 7, 8) = 8
P(L) = 8/8 = 1
```

Question-13 :-  A die 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 :-
```  The possible outcomes when a dice is thrown = {1, 2, 3, 4, 5, 6}
Number of possible outcomes of a dice = 6

(i) Prime numbers on a dice are 2, 3, and 5.
Total prime numbers on a dice = 3
P(getting a prime number) = 3/6 = 1/2

(ii) Numbers lying between 2 and 6 = 3, 4, 5
Total numbers lying between 2 and 6 = 3
P(getting a number lying between 2 and 6) = 3/6 = 1/2

(iii) Odd numbers on a dice = 1, 3, and 5
Total odd numbers on a dice = 3
P(getting an odd number) = 3/6 = 1/2
```

Question-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 cards in a well-shuffled deck = 52

(i) Total number of kings of red colour = 2
P(getting a king of red colour) = 2/52 = 1/13

(ii) Total number of face cards = 12
P(getting a face card) = 12/52 = 3/13

(iii) Total number of red face cards = 6
P(getting a red face card) = 6/52 = 3/26

(iv) Total number of Jack of hearts = 1
P(getting a Jack of hearts) = 1/52

(v) Total number of spade cards = 13
P(getting a spade card) = 13/52 = 1/4

(vi) Total number of queen of diamonds = 1
P(getting a queen of diamond) = 1/52
```

Question-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 :-
```(i) Total number of cards = 5
Total number of queens = 1
P(getting a queen) = 1/5
```
```(ii) When the queen is drawn and put aside, the total number of remaining cards will be 4.
(a) Total number of aces = 1
P(getting an ace) = 1/4
(b) As queen is already drawn, therefore, the number of queens will be 0.
P(getting a queen) = 0/4	= 0
```

Question-16 :-  12 defective pens are accidentally 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 :-
```  No. of defective pens = 12
No. of good pens = 132
Total number of pens = 12 + 132 = 144
Total number of good pens = 132

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

Question-17 :-  (i) A lot of 20 bulbs contain 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 :-
```  No. of Defective bulbs = 4
No. of Non defective bulbs = 20 - 4 = 16
Total number of bulbs = 20

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

(ii) Remaining total number of bulbs = 19
Remaining total number of non-defective bulbs = 16 − 1 = 15
P (getting a not defective bulb) = 15/19
```

Question-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 discs = 90

(i) Total number of two-digit numbers between 1 and 90 = 81
P(getting a two-digit number) = 81/90 = 9/10

(ii) Perfect squares between 1 and 90 are 1, 4, 9, 16, 25, 36, 49, 64, and 81.
Therefore, total number of perfect squares between 1 and 90 is 9.
P(getting a perfect square) = 9/90 = 1/10

(iii) Numbers that are between 1 and 90 and divisible by 5 are 5, 10, 15, 20, 25, 30,35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, and 90.
Therefore, total numbers divisible by 5 = 18
Probability of getting a number divisible by 5 = 18/90 = 1/5
```

Question-19 :-  A child has a die whose six faces show the letters as given below: The die is thrown once. What is the probability of getting (i) A? (ii) D?

Solution :-
```  Total number of possible outcomes on the dice = 6

(i) Total number of faces having A on it = 2
P(getting A) = 2/6 = 1/3

(ii) Total number of faces having D on it = 1
P(getting D) = 1/6
```

Question-20 :-  Suppose you drop a die at random on the rectangular region shown in Fig. What is the probability that it will land inside the circle with diameter 1m? Solution :-
```  Area of rectangle = l × b = 3 × 2 = 6 m²
Area of circle (of diameter 1 m) = πr² = π x (1/2)² = π/4
P(die will land inside the circle) = (π/4)/6 = π/24
```

Question-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 ?

Solution :-
```  Total number of defective pens = 20
Total number of good pens = 144 − 20 = 124
Total number of pens = 144

(i) P(getting a good pen) = 124/144 = 31/36
So, P(Nuri buys a pen) = 31/36

(ii) P(Nuri will not buy a pen) = 1 - 31/36 = 5/36
```

Question-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 1/11 .

Solution :-
```(i) It can be observed that,
To get the sum as 2, possible outcomes = (1, 1) = P(E) = 1/36
To get the sum as 3, possible outcomes = (2, 1) and (1, 2) = P(F) = 2/36 = 1/18
To get the sum as 4, possible outcomes = (3, 1), (1, 3), (2, 2) = P(G) = 3/36 = 1/12
To get the sum as 5, possible outcomes = (4, 1), (1, 4), (2, 3), (3, 2) = P(H) = 4/36 = 1/9
To get the sum as 6, possible outcomes = (5, 1), (1, 5), (2, 4), (4, 2), (3, 3) = P(I) = 5/36
To get the sum as 7, possible outcomes = (6, 1), (1, 6), (2, 5), (5, 2), (3, 4), (4, 3) = P(J) = 6/36 = 1/6
To get the sum as 8, possible outcomes = (6, 2), (2, 6), (3, 5), (5, 3), (4, 4) = P(K) = 5/36
To get the sum as 9, possible outcomes = (3, 6), (6, 3), (4, 5), (5, 4) = P(L) = 4/36 = 1/9
To get the sum as 10, possible outcomes = (4, 6), (6, 4), (5, 5) = P(M) = 3/36 = 1/12
To get the sum as 11, possible outcomes = (5, 6), (6, 5) = P(N) = 2/36 = 1/18
To get the sum as 12, possible outcomes = (6, 6) = P(O) = 1/36

(ii) Probability of each of these sums will not be 1/11 as these sums are not equally likely.
```

Question-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 possible outcomes are {HHH, TTT, HHT, HTH, THH, TTH, THT, HTT}
Number of total possible outcomes = 8
Number of favourable outcomes = 2 {i.e., TTT and HHH}

P(Hanif will win the game) = 2/8 = 1/4
P(Hanif will lose the game) = 1 - 1/4 = 3/4
```

Question-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 :-
```
Total number of outcomes = 6 × 6 = 36

(i)Total number of outcomes when 5 comes up on either time are
(5, 1), (5, 2), (5,3), (5, 4), (5, 5), (5, 6), (1, 5), (2, 5), (3, 5), (4, 5), (6, 5)
Hence, total number of favourable cases = 11
P(5 will come up either time) = 11/36
P (5 will not come up either time) = 1 - 11/36 = 25/36

(ii) Total number of cases, when 5 can come at least once = 11
P(5 will come at least once) = 11/36
```

Question-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, When two coins are tossed, the possible outcomes are (H, H), (H, T), (T, H), and (T,T).
It can be observed that there can be one of each in two possible ways − (H, T), (T, H).
Therefore, the probability of getting two heads is 1/4, the probability of getting two
tails is 1/4, and the probability of getting one of each is 1/2.
It can be observed that for each outcome, the probability is not 1/3.
```
```(ii) Correct,
When a dice is thrown, the possible outcomes are 1, 2, 3, 4, 5, and 6. Out of these,
1, 3, 5 are odd and 2, 4, 6 are even numbers.
Therefore, the probability of getting an odd number is 1/2.
```
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