NCERT Solutions for Class 10 Maths Chapter 14 Probability Exercise 14.1#
Exercise 14.1 covers classical probability — finding the probability of events using coins, dice, cards, and balls in bags.
Classical Probability:
$$P(E) = \frac{\text{Number of outcomes favourable to } E}{\text{Total number of equally likely outcomes}}$$
- $0 \leq P(E) \leq 1$
- $P(E) + P(\bar{E}) = 1$
- A standard deck has 52 cards: 4 suits (Hearts, Diamonds, Clubs, Spades), 13 each.
1. Complete the following statements: #
(i) Probability of an event E + Probability of the event ’not E’ = ___.
$P(E) + P(\bar{E}) = \mathbf{1}$
(ii) The probability of an event that cannot happen is ___. Such an event is called ___.
The probability is 0. Such an event is called an impossible event.
(iii) The probability of an event that is certain to happen is ___. Such an event is called ___.
The probability is 1. Such an event is called a sure (certain) event.
(iv) The sum of the probabilities of all the elementary events of an experiment is ___.
The sum = 1.
(v) The probability of an event is greater than or equal to ___ and less than or equal to ___.
Greater than or equal to 0 and less than or equal to 1.
2. Which of the following experiments have equally likely outcomes? Explain. #
Answer
(i) A driver attempts to start a car. The car starts or does not start. — Not equally likely (depends on car condition).
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot. — Not equally likely (depends on skill level).
(iii) A trial is made to answer a true-false question. The answer is right or wrong. — Equally likely (if guessing, both have probability 1/2).
(iv) A baby is born. It is a boy or a girl. — Equally likely (approximately 1/2 each).
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? #
Answer
Tossing a coin is considered fair because the two outcomes — heads and tails — are equally likely, each with probability $\dfrac{1}{2}$. Neither team gets an advantage, as the coin has no bias (assuming a fair coin).
4. Which of the following cannot be the probability of an event? #
Answer
(A) $\dfrac{2}{3}$ — Can be (between 0 and 1)
(B) $-1.5$ — Cannot be (negative, probability must be ≥ 0)
(C) $15\%$ = $0.15$ — Can be (between 0 and 1)
(D) $0.7$ — Can be (between 0 and 1)
Answer: (B) -1.5 cannot be the probability of an event.
5. If P(E) = 0.05, what is the probability of ’not E’? #
Answer
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?
Since the bag contains only lemon flavoured candies, taking out an orange flavoured candy is an impossible event.
$$P(\text{orange}) = \mathbf{0}$$(ii) a lemon flavoured candy?
Since the bag contains only lemon flavoured candies, this is a certain event.
$$P(\text{lemon}) = \mathbf{1}$$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? #
Answer
$P(\text{same birthday}) = 1 - P(\text{not same birthday}) = 1 - 0.992 = \mathbf{0.008}$
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?
Total balls = 8, red balls = 3.
$$P(\text{red}) = \frac{3}{8}$$(ii) not red?
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?
Total = 17.
$$P(\text{red}) = \frac{5}{17}$$(ii) white?
(iii) not green?
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?
Total = 100+50+20+10 = 180.
$$P(50p) = \frac{100}{180} = \frac{5}{9}$$(ii) will not be a ₹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 fish and 8 female fish. What is the probability that the fish taken out is a male fish? #
Answer
Total = 13, male = 5.
$$P(\text{male}) = \frac{5}{13}$$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 and these are equally likely outcomes. What is the probability that it will point at: #
(i) 8?
(ii) an odd number?
Odd numbers: 1, 3, 5, 7 → 4 numbers.
$$P(\text{odd}) = \frac{4}{8} = \frac{1}{2}$$(iii) a number greater than 2?
Numbers > 2: 3, 4, 5, 6, 7, 8 → 6 numbers.
$$P(>2) = \frac{6}{8} = \frac{3}{4}$$(iv) a number less than 9?
All 8 numbers are less than 9. This is a certain event.
$$P(<9) = \frac{8}{8} = 1$$13. A die is thrown once. Find the probability of getting: #
(i) a prime number
Prime numbers on die: 2, 3, 5 → 3 outcomes.
$$P(\text{prime}) = \frac{3}{6} = \frac{1}{2}$$(ii) a number lying between 2 and 6
Numbers between 2 and 6 (exclusive): 3, 4, 5 → 3 outcomes.
$$P(2 < x < 6) = \frac{3}{6} = \frac{1}{2}$$(iii) an odd number
Odd numbers: 1, 3, 5 → 3 outcomes.
$$P(\text{odd}) = \frac{3}{6} = \frac{1}{2}$$14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting: #
(i) a king of red colour
Red kings: King of Hearts + King of Diamonds = 2 cards.
$$P = \frac{2}{52} = \frac{1}{26}$$(ii) a face card
Face cards: Jack, Queen, King (4 suits × 3 = 12 cards).
$$P = \frac{12}{52} = \frac{3}{13}$$(iii) a red face card
Red face cards (Hearts+Diamonds, J/Q/K each) = 6 cards.
$$P = \frac{6}{52} = \frac{3}{26}$$(iv) the jack of hearts
(v) a spade
Spades = 13 cards.
$$P = \frac{13}{52} = \frac{1}{4}$$(vi) the queen of diamonds
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?
Total = 5 cards.
$$P(\text{queen}) = \frac{1}{5}$$(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?
After removing queen, 4 cards remain.
(a) an ace: $P = \dfrac{1}{4}$
(b) a queen: $P = \dfrac{0}{4} = 0$ (queen is already removed)
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. #
Answer
Total = 144, good = 132.
$$P(\text{good}) = \frac{132}{144} = \frac{11}{12}$$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? #
(i) Probability of defective bulb
Total = 20, defective = 4.
$$P(\text{defective}) = \frac{4}{20} = \frac{1}{5}$$(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?
After removing one good bulb: 19 bulbs remain, 4 defective, 15 good.
$$P(\text{not defective}) = \frac{15}{19}$$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
Two-digit numbers from 10 to 90 = 81 numbers.
$$P = \frac{81}{90} = \frac{9}{10}$$(ii) a perfect square number
Perfect squares from 1 to 90: 1, 4, 9, 16, 25, 36, 49, 64, 81 → 9 numbers.
$$P = \frac{9}{90} = \frac{1}{10}$$(iii) a number divisible by 5
Multiples of 5: 5, 10, 15, …, 90 → 18 numbers.
$$P = \frac{18}{90} = \frac{1}{5}$$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? #
(i) A
A appears on 2 faces out of 6.
$$P(A) = \frac{2}{6} = \frac{1}{3}$$(ii) D
D appears on 1 face.
$$P(D) = \frac{1}{6}$$20. Suppose you drop a die at random on the rectangular region shown in the figure. What is the probability that it will land inside the circle with diameter 1 m? #
Answer
Rectangle: $3 \times 2 = 6$ m². Circle: radius = 0.5 m, area = $\pi(0.5)^2 = 0.25\pi$ m².
$$P = \frac{0.25\pi}{6} = \frac{\pi}{24} \approx \frac{3.14}{24} \approx 0.131$$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?
Good pens = 124.
$$P(\text{buy}) = \frac{124}{144} = \frac{31}{36}$$(ii) She will not buy it?
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. Do you agree with this argument? Justify your answer. #
Answer
When two dice are thrown, total outcomes = 36.
Sum table:
| Sum | Outcomes | Probability |
|---|---|---|
| 2 | (1,1) | 1/36 |
| 3 | (1,2),(2,1) | 2/36 |
| 4 | (1,3),(2,2),(3,1) | 3/36 |
| 5 | (1,4),(2,3),(3,2),(4,1) | 4/36 |
| 6 | (1,5),(2,4),(3,3),(4,2),(5,1) | 5/36 |
| 7 | (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) | 6/36 |
| 8 | (2,6),(3,5),(4,4),(5,3),(6,2) | 5/36 |
| 9 | (3,6),(4,5),(5,4),(6,3) | 4/36 |
| 10 | (4,6),(5,5),(6,4) | 3/36 |
| 11 | (5,6),(6,5) | 2/36 |
| 12 | (6,6) | 1/36 |
No, the student is not correct. The 11 outcomes are not equally likely. For example, sum 7 can occur in 6 ways while sum 2 can occur in only 1 way. Therefore, each sum does not have probability 1/11.
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. #
Answer
Total outcomes when tossing 3 coins = 8: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.
Winning outcomes (all same): HHH, TTT → 2.
$$P(\text{win}) = \frac{2}{8} = \frac{1}{4}$$$$P(\text{lose}) = 1 - \frac{1}{4} = \frac{3}{4}$$24. A die is thrown twice. What is the probability that: #
(i) 5 will not come up either time?
P(5 on one throw) = 1/6, P(not 5) = 5/6.
$$P(\text{5 not on both}) = \frac{5}{6} \times \frac{5}{6} = \frac{25}{36}$$(ii) 5 will come up at least once?
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.
Not correct. The three outcomes are not equally likely. The sample space is {HH, HT, TH, TT} with 4 equally likely outcomes. $P(\text{HH}) = P(\text{TT}) = \dfrac{1}{4}$ and $P(\text{one of each}) = \dfrac{2}{4} = \dfrac{1}{2}$.
(ii) If a die is thrown, there are two possible outcomes — odd number or even number. Therefore, the probability of getting an odd number is 1/2.
Correct. The die has 3 odd numbers {1,3,5} and 3 even numbers {2,4,6}, each with probability $\dfrac{3}{6} = \dfrac{1}{2}$. The two outcomes “odd” and “even” are equally likely.