A game consists of tossing a one rupee coin 3 times. Hanif wins if all the tosses give the same result. Calculate the probability that Hanif will lose the game.

Total outcomes = 8. Favourable outcomes for losing (not same result) = 6. Probability = 6/8 = 3/4.

A die is thrown twice. What is the probability that 5 will come up at least once?

Favourable outcomes = 11. Total outcomes = 36. Probability = 11/36.

NCERT Solutions Class 10 Maths Chapter 14: Probability

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

(B) -1.5. Probability cannot be negative.

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

P(not E) = 1 - P(E) = 1 - 0.05 = 0.95.

Q: 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.

(i) Prime number (2, 3, 5): 3/6 = 1/2. (ii) Between 2 and 6 (3, 4, 5): 3/6 = 1/2. (iii) Odd number (1, 3, 5): 3/6 = 1/2.

Q: 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

(i) King of red colour: 2/52 = 1/26. (ii) Face card: 12/52 = 3/13.

Detailed solutions for Exercise 14.1.

Ex 14.1

#### **Q1. 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 __________ .** #### **A1. Solution:** (i) **1** (ii) **0**, **Impossible event** (iii) **1**, **Sure event** or **Certain event** (iv) **1** (v) **0**, **1** #### **Q2. 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.** #### **A2. Solution:** **(i) Not equally likely:** The car starting depends on various factors (fuel, engine condition, etc.), not just chance. **(ii) Not equally likely:** It depends on the player's skill. **(iii) Equally likely:** There are only two possibilities (True or False) and both are fair possibilities. **(iv) Equally likely:** There are only two possibilities (Boy or Girl) and they are generally considered equally likely. #### **Q3. 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?** #### **A3. Solution:** Tossing a coin is considered fair because the outcomes (Head and Tail) are **equally likely**. The result of an individual toss is completely unpredictable. #### **Q4. Which of the following cannot be the probability of an event?** **(A) $\frac{2}{3}$ (B) $-1.5$ (C) $15\%$ (D) $0.7$** #### **A4. Solution:** **(B) $-1.5$**. Probability cannot be negative. It must be between 0 and 1. #### **Q5. If $P(E) = 0.05$, what is the probability of ‘not E’?** #### **A5. Solution:** $P(\text{not E}) = 1 - P(E)$ $= 1 - 0.05 = 0.95$. #### **Q6. 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?** #### **A6. Solution:** **(i) 0:** The event is impossible as there are only lemon candies. **(ii) 1:** The event is sure as all candies are lemon flavoured. #### **Q7. 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?** #### **A7. Solution:** $P(\text{same birthday}) + P(\text{not same birthday}) = 1$ $P(\text{same birthday}) = 1 - 0.992 = 0.008$. #### **Q8. 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?** #### **A8. Solution:** Total balls = $3 + 5 = 8$. **(i) P(Red):** $\frac{\text{Number of red balls}}{\text{Total balls}} = \frac{3}{8}$. **(ii) P(Not Red):** $1 - P(\text{Red}) = 1 - \frac{3}{8} = \frac{5}{8}$. #### **Q9. 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?** #### **A9. Solution:** Total marbles = $5 + 8 + 4 = 17$. **(i) P(Red):** $\frac{5}{17}$. **(ii) P(White):** $\frac{8}{17}$. **(iii) P(Not Green):** $\frac{\text{Red} + \text{White}}{\text{Total}} = \frac{5 + 8}{17} = \frac{13}{17}$. #### **Q10. 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 50p coin? (ii) will not be a ₹ 5 coin?** #### **A10. Solution:** Total coins = $100 + 50 + 20 + 10 = 180$. **(i) P(50p coin):** $\frac{100}{180} = \frac{5}{9}$. **(ii) P(Not ₹ 5 coin):** Total non-₹5 coins = $100 + 50 + 20 = 170$. $P = \frac{170}{180} = \frac{17}{18}$. #### **Q11. 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?** #### **A11. Solution:** Total fish = $5 + 8 = 13$. $P(\text{Male fish}) = \frac{5}{13}$. #### **Q12. 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?** **(iii) a number greater than 2?** **(iv) a number less than 9?** #### **A12. Solution:** Total outcomes = 8. **(i) P(8):** $\frac{1}{8}$. **(ii) P(Odd):** Odd numbers are 1, 3, 5, 7 (4 numbers). $P = \frac{4}{8} = \frac{1}{2}$. **(iii) P(> 2):** Numbers are 3, 4, 5, 6, 7, 8 (6 numbers). $P = \frac{6}{8} = \frac{3}{4}$. **(iv) P(< 9):** All numbers are less than 9. $P = \frac{8}{8} = 1$. #### **Q13. 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.** #### **A13. Solution:** Total outcomes = 6 (1, 2, 3, 4, 5, 6). **(i) Prime number:** 2, 3, 5 (3 numbers). $P = \frac{3}{6} = \frac{1}{2}$. **(ii) Between 2 and 6:** 3, 4, 5 (3 numbers). $P = \frac{3}{6} = \frac{1}{2}$. **(iii) Odd number:** 1, 3, 5 (3 numbers). $P = \frac{3}{6} = \frac{1}{2}$. #### **Q14. 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** #### **A14. Solution:** Total cards = 52. **(i) King of red colour:** (King of Hearts, King of Diamonds) = 2. $P = \frac{2}{52} = \frac{1}{26}$. **(ii) Face card:** (J, Q, K in 4 suits) = 12. $P = \frac{12}{52} = \frac{3}{13}$. **(iii) Red face card:** (J, Q, K in Hearts and Diamonds) = 6. $P = \frac{6}{52} = \frac{3}{26}$. **(iv) Jack of hearts:** 1 card. $P = \frac{1}{52}$. **(v) A spade:** 13 cards. $P = \frac{13}{52} = \frac{1}{4}$. **(vi) Queen of diamonds:** 1 card. $P = \frac{1}{52}$. #### **Q15. 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?** #### **A15. Solution:** Total cards = 5. **(i) P(Queen):** $\frac{1}{5}$. **(ii) Queen is put aside:** Total remaining = 4. **(a) P(Ace):** $\frac{1}{4}$. **(b) P(Queen):** 0 (since the queen was removed). #### **Q16. 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.** #### **A16. Solution:** Total pens = $12 + 132 = 144$. Good pens = 132. $P(\text{Good pen}) = \frac{132}{144} = \frac{11}{12}$. #### **Q17. (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?** #### **A17. Solution:** **(i)** Total = 20, Defective = 4. $P(\text{Defective}) = \frac{4}{20} = \frac{1}{5}$. **(ii)** Remaining total = 19. Remaining good bulbs = $16 - 1 = 15$. $P(\text{Not defective}) = \frac{15}{19}$. #### **Q18. 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.** #### **A18. Solution:** Total outcomes = 90. **(i) Two-digit number:** Numbers from 10 to 90. Count = $90 - 9 = 81$. $P = \frac{81}{90} = \frac{9}{10}$. **(ii) Perfect square:** 1, 4, 9, 16, 25, 36, 49, 64, 81 (9 numbers). $P = \frac{9}{90} = \frac{1}{10}$. **(iii) Divisible by 5:** 5, 10, ..., 90. Count = 18. $P = \frac{18}{90} = \frac{1}{5}$. #### **Q19. 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?** #### **A19. Solution:** Total faces = 6. **(i) P(A):** 'A' appears 2 times. $P = \frac{2}{6} = \frac{1}{3}$. **(ii) P(D):** 'D' appears 1 time. $P = \frac{1}{6}$. #### **Q20. 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?** #### **A20. Solution:** Total pens = 144. Defective = 20. Good = 124. **(i) P(She buys):** She buys if it is good. $P = \frac{124}{144} = \frac{31}{36}$. **(ii) P(She will not buy):** She doesn't buy if defective. $P = \frac{20}{144} = \frac{5}{36}$. #### **Q21. Refer to Example 13. (Two dice are thrown at the same time).** **(i) Complete the following table:** *(Sum of two dice vs Probability)* **(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.** #### **A21. Solution:** Total outcomes when two dice are thrown = $6 \times 6 = 36$. **(i)** - Sum 2: (1,1) $\rightarrow 1/36$ - Sum 3: (1,2), (2,1) $\rightarrow 2/36$ - Sum 4: (1,3), (2,2), (3,1) $\rightarrow 3/36$ - Sum 5: (1,4), (2,3), (3,2), (4,1) $\rightarrow 4/36$ - Sum 6: (1,5), (2,4), (3,3), (4,2), (5,1) $\rightarrow 5/36$ - Sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) $\rightarrow 6/36$ - Sum 8: (2,6), (3,5), (4,4), (5,3), (6,2) $\rightarrow 5/36$ - Sum 9: (3,6), (4,5), (5,4), (6,3) $\rightarrow 4/36$ - Sum 10: (4,6), (5,5), (6,4) $\rightarrow 3/36$ - Sum 11: (5,6), (6,5) $\rightarrow 2/36$ - Sum 12: (6,6) $\rightarrow 1/36$ **(ii) No.** The sums are not equally likely. For example, Sum 7 has 6 favourable outcomes, while Sum 2 has only 1. #### **Q22. 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.** #### **A22. Solution:** Total outcomes = $2^3 = 8$ (HHH, HHT, HTH, THH, HTT, THT, TTH, TTT). Favourable outcomes for winning (Same result): HHH, TTT (2 outcomes). Favourable outcomes for losing: $8 - 2 = 6$. $P(\text{Losing}) = \frac{6}{8} = \frac{3}{4}$. #### **Q23. 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?** #### **A23. Solution:** Total outcomes = 36. **(ii) 5 comes up at least once:** (1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (5,1), (5,2), (5,3), (5,4), (5,6). Total = 11. $P(\text{5 at least once}) = \frac{11}{36}$. **(i) 5 will not come up either time:** $P = 1 - P(\text{5 at least once}) = 1 - \frac{11}{36} = \frac{25}{36}$. #### **Q24. 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.** #### **A24. Solution:** **(i) Incorrect.** The outcomes are HH, HT, TH, TT. "One of each" includes HT and TH, so its probability is $2/4 = 1/2$, not $1/3$. **(ii) Correct.** Odd numbers (1,3,5) and Even numbers (2,4,6) are equal in number (3 each). So, probability is $3/6 = 1/2$.