Class 10 NCERT Solutions
Chapter 14: Probability
Master the Basic Proportionality Theorem (BPT), the criteria for similarity (SAS, SSS, AA), and the logical proofs of triangles with our step-by-step breakdown.
Exercise 14.1
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 _____.
(i) 1
(ii) 0; impossible event
(iii) 1; sure (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. (ii) A player shoots a basketball. (iii) Answering a true-false question. (iv) A baby is born.
(i) Not equally likely. Whether the car starts depends on many external factors (fuel, battery condition, etc.) that are not equally balanced.
(ii) Not equally likely. The outcome depends on the player’s skill and practice level, so the two outcomes are not equally probable.
(iii) Equally likely. The answer is either right or wrong with equal chance, assuming the respondent has no knowledge of the topic.
(iv) Equally likely. A baby being born as a boy or girl is considered equally likely in probability theory.
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?
When we toss a coin, there are only two possible outcomes — head or tail — and both are equally likely. Since the result of each individual toss is completely unpredictable, neither team has an advantage. This makes it a fair and unbiased method of decision-making.
4. Which of the following cannot be the probability of an event?
(A) $\frac{2}{3}$ (B) $-1.5$ (C) $15\%$ (D) $0.7$
The probability of any event must always satisfy $0 \leq P(E) \leq 1$. It can never be negative or greater than 1.
Since $-1.5 < 0$, it cannot be a probability.
Answer: (B)
5. If $P(E) = 0.05$, what is the probability of “not E”?
Using the complement rule:
$$P(\overline{E}) = 1-P(E) = 1-0.05 = \mathbf{0.95}$$
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?
(i) Since the bag contains only lemon flavoured candies, picking an orange flavoured candy is an impossible event.
$$P(\text{orange flavoured candy}) = 0$$(ii) Every candy in the bag is lemon flavoured, so picking a lemon flavoured candy is a certain (sure) event.
$$P(\text{lemon flavoured candy}) = 1$$
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?
8. A bag contains 3 red balls and 5 black balls. A ball is drawn at random. What is the probability that the ball drawn is (i) red? (ii) not red?
Total balls $= 8$.
(i) $P(\text{red}) = \dfrac{3}{8}$
(ii) $P(\text{not red}) = 1-\dfrac{3}{8} = \dfrac{5}{8}$
9. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out at random. What is the probability that the marble taken out will be (i) red? (ii) white? (iii) not green?
Total marbles $= 5+8+4 = 17$.
(i) $P(\text{red}) = \dfrac{5}{17}$
(ii) $P(\text{white}) = \dfrac{8}{17}$
(iii) $P(\text{not green}) = 1-\dfrac{4}{17} = \dfrac{13}{17}$
10. A piggy bank contains 100 fifty-paise coins, 50 Rs 1 coins, 20 Rs 2 coins and 10 Rs 5 coins. What is the probability that the coin that falls out will (i) be a 50 p coin? (ii) not be a Rs 5 coin?
Total coins $= 100+50+20+10 = 180$.
(i) $P(\text{50 p coin}) = \dfrac{100}{180} = \dfrac{5}{9}$
(ii) $P(\text{not Rs 5 coin}) = 1-\dfrac{10}{180} = 1-\dfrac{1}{18} = \dfrac{17}{18}$
11. A 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?
Total fishes $= 5+8 = 13$.
$$P(\text{male fish}) = \frac{5}{13}$$
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 (equally likely). 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 outcomes $= 8$.
(i) $P(8) = \dfrac{1}{8}$
(ii) Odd numbers: 1, 3, 5, 7 → $P = \dfrac{4}{8} = \dfrac{1}{2}$
(iii) Numbers greater than 2: 3, 4, 5, 6, 7, 8 → $P = \dfrac{6}{8} = \dfrac{3}{4}$
(iv) All 8 numbers are less than 9 → $P = \dfrac{8}{8} = 1$
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.
Total outcomes $= \{1,2,3,4,5,6\}$, so 6 outcomes.
(i) Prime numbers: 2, 3, 5 → $P = \dfrac{3}{6} = \dfrac{1}{2}$
(ii) Numbers between 2 and 6 (exclusive): 3, 4, 5 → $P = \dfrac{3}{6} = \dfrac{1}{2}$
(iii) Odd numbers: 1, 3, 5 → $P = \dfrac{3}{6} = \dfrac{1}{2}$
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 (v) a spade (vi) the queen of diamonds
(i) $P(\text{red king}) = \dfrac{2}{52} = \dfrac{1}{26}$
(ii) $P(\text{face card}) = \dfrac{12}{52} = \dfrac{3}{13}$
(iii) $P(\text{red face card}) = \dfrac{6}{52} = \dfrac{3}{26}$
(iv) $P(\text{jack of hearts}) = \dfrac{1}{52}$
(v) $P(\text{spade}) = \dfrac{13}{52} = \dfrac{1}{4}$
(vi) $P(\text{queen of diamonds}) = \dfrac{1}{52}$
15. Five cards — the ten, jack, queen, king and ace of diamonds — are well-shuffled 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) Total $= 5$ cards. $P(\text{queen}) = \dfrac{1}{5}$
(ii) After removing the queen, 4 cards remain.
(a) $P(\text{ace}) = \dfrac{1}{4}$
(b) There are no queens left. $P(\text{queen}) = \dfrac{0}{4} = 0$
16. 12 defective pens are accidentally mixed with 132 good ones. One pen is taken out at random. Determine the probability that the pen taken out is a good one.
Total pens $= 12+132 = 144$.
$$P(\text{good pen}) = \frac{132}{144} = \frac{11}{12}$$
17. (i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random. What is the probability that this bulb is defective?
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. What is the probability that the next bulb drawn is not defective?
(i) Total $= 20$, defective $= 4$. $P(\text{defective}) = \dfrac{4}{20} = \dfrac{1}{5}$
(ii) Remaining: 19 bulbs total, 15 non-defective. $P(\text{not defective}) = \dfrac{15}{19}$
18. A box contains 90 discs numbered from 1 to 90. If one disc is drawn at random, find the probability that it bears:
(i) a two-digit number (ii) a perfect square number (iii) a number divisible by 5
Total discs $= 90$.
(i) Two-digit numbers from 1 to 90: 10 to 90 → $81$ numbers. $P = \dfrac{81}{90} = \dfrac{9}{10}$
(ii) Perfect squares: 1, 4, 9, 16, 25, 36, 49, 64, 81 → $9$ numbers. $P = \dfrac{9}{90} = \dfrac{1}{10}$
(iii) Multiples of 5 from 1 to 90 → $18$ numbers. $P = \dfrac{18}{90} = \dfrac{1}{5}$
19. A child has a die whose six faces show the letters A, B, C, D, E, A. The die is thrown once. What is the probability of getting (i) A? (ii) D?
Total outcomes $= 6$.
(i) Letter A appears on 2 faces. $P(A) = \dfrac{2}{6} = \dfrac{1}{3}$
(ii) Letter D appears on 1 face. $P(D) = \dfrac{1}{6}$
20. Suppose you drop a die at random on a rectangular region of dimensions 3 m × 2 m. What is the probability that it will land inside a circle with diameter 1 m?
21. A lot consists of 144 ball pens of which 20 are defective. The shopkeeper draws one pen at random.
(i) What is the probability that she will buy it?
(ii) What is the probability that she will not buy it?
Good pens $= 144-20 = 124$.
(i) Nuri buys only good pens. $P(\text{buy}) = \dfrac{124}{144} = \dfrac{31}{36}$
(ii) $P(\text{not buy}) = 1-\dfrac{31}{36} = \dfrac{5}{36}$
22. Two dice, one blue and one grey, are thrown at the same time.
(i) Write down all possible outcomes and complete the probability table for sums 2 through 12.
(ii) A student argues that each of the 11 possible sums has probability 1/11. Do you agree?
(i) By listing all 36 equally likely outcomes:
| Sum | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Probability | $\frac{1}{36}$ | $\frac{2}{36}$ | $\frac{3}{36}$ | $\frac{4}{36}$ | $\frac{5}{36}$ | $\frac{6}{36}$ | $\frac{5}{36}$ | $\frac{4}{36}$ | $\frac{3}{36}$ | $\frac{2}{36}$ | $\frac{1}{36}$ |
(ii) No, the student is incorrect. The 11 sums are not equally likely — they arise from different numbers of outcome combinations. For example, there is only 1 way to get sum 2 but 6 ways to get sum 7.
23. A game consists of tossing a coin 3 times. Hanif wins if all the tosses give the same result. Calculate the probability that Hanif will lose the game.
Possible outcomes: $\{$HHH, TTT, HHT, HTH, THH, TTH, THT, HTT$\}$ → 8 total.
Favourable outcomes for Hanif (all same): HHH, TTT → 2 outcomes.
$$P(\text{Hanif wins}) = \frac{2}{8} = \frac{1}{4}$$ $$P(\text{Hanif loses}) = 1-\frac{1}{4} = \frac{3}{4}$$
24. A die is thrown twice. What is the probability that (i) 5 will not come up either time? (ii) 5 will come up at least once?
Total outcomes $= 36$.
Outcomes where 5 appears at least once: $(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(1,5),(2,5),(3,5),(4,5),(6,5)$ → 11 outcomes.
(i) $P(\text{5 not up either time}) = 1-\dfrac{11}{36} = \dfrac{25}{36}$
(ii) $P(\text{5 at least once}) = \dfrac{11}{36}$
25. Which of the following arguments are correct and which are not correct? Give reasons.
(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.
(i) Incorrect. When two coins are tossed, there are 4 equally likely outcomes: (H,H), (H,T), (T,H), (T,T). Getting one head and one tail can happen in 2 ways. So the correct probabilities are: $P(\text{two heads}) = \dfrac{1}{4}$, $P(\text{two tails}) = \dfrac{1}{4}$, $P(\text{one of each}) = \dfrac{2}{4} = \dfrac{1}{2}$. The three outcomes are not equally likely, so $\dfrac{1}{3}$ is wrong.
(ii) Correct. The outcomes $\{1,2,3,4,5,6\}$ split evenly: $\{1,3,5\}$ are odd and $\{2,4,6\}$ are even. Each group has 3 outcomes out of 6, giving $P(\text{odd}) = \dfrac{3}{6} = \dfrac{1}{2}$.
