NCERT Solutions For Class 10 Maths: Probability

EXERCISE 14.1

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-
1.(i) 1.

2.(ii) 0, impossible event.

3.(iii) 1, sure event or certain event.

4.(iv) 1.

5.(v) 0, 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.
(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.

Solution-
1.(i) Not equally likely. This is because the car may be more likely to start or not start depending on the condition of the car, introducing bias.

2.(ii) Not equally likely. The outcome depends on the skill of the player, meaning the probability of shooting or missing is not necessarily equal.

3.(iii) Equally likely. Assuming the question is fair and the answer is guessed, there is no bias, and the two possible outcomes (right or wrong) are symmetrical, meaning each is as likely to occur as the other.

4.(iv) Equally likely. The assumption is that the birth outcomes (boy or girl) are symmetrical.

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-
1.Tossing a coin is considered a fair way because the outcomes of a random toss of an unbiased or fair coin (Head or Tail) are equally likely.

2.There is no reason for the coin to land more often on one side than the other, guaranteeing a decision without any bias or interference.

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-
1.The probability $P(E)$ of an event E must satisfy the condition $0 \leq P(E) \leq 1$.

2.Option (B) is $-1.5$. Since probability cannot be negative, $-1.5$ cannot be the probability of an event.

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

Solution-
1.The event ‘not E’, denoted by $\bar{E}$, is the complement of event E.
2.We use the relationship: $P(\bar{E}) = 1 - P(E)$.
3.$P(\text{not E}) = 1 - 0.05$
4.$P(\text{not E}) = 0.95$.

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-
1.(i) Probability of taking out an orange flavoured candy:
Since the bag contains only lemon flavoured candies, drawing an orange flavoured candy is an impossible event.
$P(\text{Orange flavoured candy}) = 0$.

2.(ii) Probability of taking out a lemon flavoured candy:
Since all candies are lemon flavoured, drawing a lemon flavoured candy is a sure event or certain event.
$P(\text{Lemon flavoured candy}) = 1$.

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-
1.Let E be the event that the 2 students have the same birthday. Let $\bar{E}$ be the event that the 2 students not having the same birthday (0.992). These are complementary events.
2.We use the formula $P(E) = 1 - P(\bar{E})$.
3.$P(\text{Same birthday}) = 1 - 0.992$
4.$P(\text{Same birthday}) = 0.008$.

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-
1.Total number of possible outcomes $= 3 + 5 = 8$.

2.(i) Probability that the ball drawn is red:
$P(\text{Red}) = \frac{\text{Number of outcomes favourable to Red}}{\text{Number of all possible outcomes}} = \frac{3}{8}$.

3.(ii) Probability that the ball drawn is not red:
The event 'not red' is the complement of 'red'.
$P(\text{not Red}) = 1 - P(\text{Red}) = 1 - \frac{3}{8} = \frac{5}{8}$.
(Alternatively, the favourable outcomes are the 5 black balls.)

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-
1.Total number of possible outcomes $= 5 + 8 + 4 = 17$.

2.(i) Probability that the marble is red:
$P(\text{Red}) = \frac{5}{17}$.

3.(ii) Probability that the marble is white:
$P(\text{White}) = \frac{8}{17}$.

4.(iii) Probability that the marble is not green:
The number of marbles that are not green is $5 + 8 = 13$.
$P(\text{not Green}) = \frac{13}{17}$.
(Alternatively, $P(\text{not Green}) = 1 - P(\text{Green}) = 1 - 4/17 = 13/17$.)

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-
1.Total number of coins (possible outcomes) $= 100 + 50 + 20 + 10 = 180$.

2.(i) Probability that the coin will be a 50 p coin:
Number of favourable outcomes $= 100$.
$P(\text{50 p coin}) = \frac{100}{180} = \frac{5}{9}$.

3.(ii) Probability that the coin will not be a ₹ 5 coin:
Number of ₹ 5 coins $= 10$.
Number of coins that are not ₹ 5 $= 180 - 10 = 170$.
$P(\text{not ₹ 5 coin}) = \frac{170}{180} = \frac{17}{18}$.

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. 14.4). What is the probability that the fish taken out is a male fish?

Solution-
1.Total number of fish (possible outcomes) $= 5 + 8 = 13$.

2.The number of outcomes favourable to the event 'fish is a male fish' is 5.
$P(\text{Male fish}) = \frac{5}{13}$.

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. 14.5 ), 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-
1.Total number of possible outcomes $= 8$.

2.(i) Probability that it will point at 8:
$P(8) = \frac{1}{8}$.

3.(ii) Probability that it will point at an odd number:
Odd numbers are {1, 3, 5, 7}. Number of favourable outcomes $= 4$.
$P(\text{Odd number}) = \frac{4}{8} = \frac{1}{2}$.

4.(iii) Probability that it will point at a number greater than 2:
Numbers greater than 2 are {3, 4, 5, 6, 7, 8}. Number of favourable outcomes $= 6$.
$P(\text{Number} > 2) = \frac{6}{8} = \frac{3}{4}$.

5.(iv) Probability that it will point at a number less than 9:
Numbers less than 9 are {1, 2, 3, 4, 5, 6, 7, 8}. Number of favourable outcomes $= 8$. This is a sure event.
$P(\text{Number} < 9) = \frac{8}{8} = 1$.

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-
1.Total number of possible outcomes when a die is thrown once is 6: {1, 2, 3, 4, 5, 6}.

2.(i) Probability of getting a prime number:
Prime numbers are {2, 3, 5}. Number of favourable outcomes $= 3$.
$P(\text{Prime number}) = \frac{3}{6} = \frac{1}{2}$.

3.(ii) Probability of getting a number lying between 2 and 6:
Numbers between 2 and 6 are {3, 4, 5}. Number of favourable outcomes $= 3$.
$P(\text{Number between 2 and 6}) = \frac{3}{6} = \frac{1}{2}$.

4.(iii) Probability of getting an odd number:
Odd numbers are {1, 3, 5}. Number of favourable outcomes $= 3$.
$P(\text{Odd number}) = \frac{3}{6} = \frac{1}{2}$.

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 (v) a spade (vi) the queen of diamonds

Solution-
1.Total number of possible outcomes $= 52$.

2.(i) Probability of getting a king of red colour:
There are 2 Kings of red colour (King of Hearts and King of Diamonds). Favourable outcomes $= 2$.
$P(\text{Red King}) = \frac{2}{52} = \frac{1}{26}$.

3.(ii) Probability of getting a face card:
There are 12 face cards in total (3 in each of the 4 suits). Favourable outcomes $= 12$.
$P(\text{Face card}) = \frac{12}{52} = \frac{3}{13}$.

4.(iii) Probability of getting a red face card:
There are 6 red face cards (3 from Hearts and 3 from Diamonds). Favourable outcomes $= 6$.
$P(\text{Red face card}) = \frac{6}{52} = \frac{3}{26}$.

5.(iv) Probability of getting the jack of hearts:
There is exactly one Jack of hearts. Favourable outcome $= 1$.
$P(\text{Jack of hearts}) = \frac{1}{52}$.

6.(v) Probability of getting a spade:
There are 13 spades. Favourable outcomes $= 13$.
$P(\text{Spade}) = \frac{13}{52} = \frac{1}{4}$.

7.(vi) Probability of getting the queen of diamonds:
There is exactly one Queen of diamonds. Favourable outcome $= 1$.
$P(\text{Queen of diamonds}) = \frac{1}{52}$.

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-
1.Initial total number of possible outcomes $= 5$.

2.(i) Probability that the card is the queen:
There is 1 Queen in the set.
$P(\text{Queen}) = \frac{1}{5}$.

3.(ii) If the queen is drawn and put aside, the total number of remaining cards is $5 - 1 = 4$.

4.(a) Probability that the second card is an ace:
There is 1 Ace remaining.
$P(\text{Ace}) = \frac{1}{4}$.

5.(b) Probability that the second card is a queen:
There are 0 Queens remaining.
$P(\text{Queen}) = \frac{0}{4} = 0$.

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-
1.Total number of pens in the lot (possible outcomes) $= 12 (\text{defective}) + 132 (\text{good}) = 144$.

2.The number of outcomes favourable to the event 'pen taken out is a good one' is 132.
$P(\text{Good pen}) = \frac{132}{144}$

3.Simplifying the fraction: $P(\text{Good pen}) = \frac{11}{12}$.

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-
1.Initial total bulbs $= 20$. Defective bulbs $= 4$. Good bulbs $= 16$.

2.(i) Probability that the bulb drawn is defective:
$P(\text{Defective}) = \frac{4}{20} = \frac{1}{5}$.

3.(ii) If the bulb drawn in (i) was not defective (i.e., good) and not replaced, the new totals are:
Total remaining bulbs $= 20 - 1 = 19$.
Good bulbs remaining $= 16 - 1 = 15$.

4.Probability that the second bulb drawn is not defective (i.e., good):
$P(\text{Not defective}) = \frac{15}{19}$.

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-
1.Total number of possible outcomes $= 90$.

2.(i) Probability that it bears a two-digit number:
Two-digit numbers range from 10 to 90. Number of favourable outcomes $= 90 - 9 = 81$.
$P(\text{Two-digit number}) = \frac{81}{90} = \frac{9}{10}$.

3.(ii) Probability that it bears a perfect square number:
Perfect square numbers from 1 to 90 are {1, 4, 9, 16, 25, 36, 49, 64, 81}. Number of favourable outcomes $= 9$.
$P(\text{Perfect square number}) = \frac{9}{90} = \frac{1}{10}$.

4.(iii) Probability that it bears a number divisible by 5:
Numbers divisible by 5 are {5, 10, 15, ..., 90}. Number of favourable outcomes $= 18$.
$P(\text{Divisible by 5}) = \frac{18}{90} = \frac{1}{5}$.

Question 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-
1.Total number of possible outcomes $= 6$.

2.(i) Probability of getting A:
The letter A appears on 2 faces. Number of favourable outcomes $= 2$.
$P(A) = \frac{2}{6} = \frac{1}{3}$.

3.(ii) Probability of getting D:
The letter D appears on 1 face. Number of favourable outcome $= 1$.
$P(D) = \frac{1}{6}$.

Question 20: Suppose you drop a die at random on the rectangular region shown in Fig. 14.6. What is the probability that it will land inside the circle with diameter 1m?

Solution-
1.This problem requires calculating geometrical probability, defined as the ratio of the favourable area to the total area.

2.Calculate the Total Area (Rectangular Region): Dimensions are $3$ m by $2$ m.
Total Area $= 3 \times 2 = 6$ m$^2$.

3.Calculate the Favourable Area (Circle): Diameter $D=1$ m, so radius $r = 1/2$ m.
Favourable Area $= \pi r^2 = \pi (1/2)^2 = \frac{\pi}{4}$ m$^2$.

4.Calculate the Probability:
$P(\text{Inside circle}) = \frac{\text{Area of circle}}{\text{Area of rectangle}} = \frac{\pi/4}{6}$

5.$P(\text{Inside circle}) = \frac{\pi}{24}$.

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-
1.Total number of pens $= 144$. Defective pens $= 20$. Good pens $= 144 - 20 = 124$.

2.(i) Probability that she will buy it (i.e., the pen is good):
Number of favourable outcomes $= 124$.
$P(\text{Buy}) = \frac{124}{144} = \frac{31}{36}$.

3.(ii) Probability that she will not buy it (i.e., the pen is defective):
Number of favourable outcomes $= 20$.
$P(\text{Not Buy}) = \frac{20}{144} = \frac{5}{36}$.

Question 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-
1.(i) The possible outcomes when throwing two dice total 36. The completed table is based on the number of combinations yielding each sum:

2.(ii) Justification:
1.The student's argument is not correct.
2.The theoretical probability definition requires that all outcomes of the experiment must be equally likely.
3.While the sums 2 through 12 are all possible outcomes, they are not equally likely because the number of ways to achieve each sum is different (e.g., Sum 7 occurs 6 times, Sum 2 occurs 1 time).

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-
1.The possible outcomes when tossing a coin 3 times are 8: {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. Total outcomes $= 8$.

2.Hanif wins if the result is HHH or TTT. (2 favourable outcomes).

3.Hanif loses otherwise. The number of outcomes favourable to losing $= 8 - 2 = 6$.

4.Calculate the probability that Hanif will lose the game:
$P(\text{Lose}) = \frac{6}{8} = \frac{3}{4}$.

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?
[Hint : Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]

Solution-
1.Total number of possible outcomes $= 36$.

2.(i) Probability that 5 will not come up either time:
If 5 does not come up, the outcome of each throw must be one of {1, 2, 3, 4, 6}.
Number of outcomes where 5 does not appear $= 5 \times 5 = 25$.
$P(\text{5 will not come up}) = \frac{25}{36}$.

3.(ii) Probability that 5 will come up at least once:
Let E be the event that '5 comes up at least once'. This is the complement of the event $\bar{E}$ where '5 will not come up either time'.
$P(E) = 1 - P(\bar{E}) = 1 - \frac{25}{36}$
$P(E) = \frac{11}{36}$.

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-
1.(i) The argument is not correct.
1.The elementary events are (H, H), (T, T), (H, T), and (T, H). Total outcomes $= 4$.
2.The events stated (two heads, two tails, one of each) are not equally likely because the outcome 'one of each' (HT or TH) has a higher frequency (2 out of 4) than the others (1 out of 4).

2.(ii) The argument is correct.
1.The elementary outcomes are {1, 2, 3, 4, 5, 6}.
2.The outcomes favorable to 'odd number' are {1, 3, 5} (3 outcomes). The outcomes favorable to 'even number' are {2, 4, 6} (3 outcomes).
3.Since the outcomes (Odd and Even) are equally likely (both having a count of 3), the probability of getting an odd number is $3/6 = 1/2$.