Chapter 14: Probability Exercise & In-Text Questions & Answers
1Complete the following statements:
Answer & Reference:
(i) Probability of an event E + Probability of the event 'not E' = 1. (Page 206, Remark)
(ii) The probability of an event that cannot happen is 0. Such an event is called impossible event. (Page 207)
(iii) The probability of an event that is certain to happen is 1. Such an event is called sure event or certain event. (Page 207)
(iv) The sum of the probabilities of all the elementary events of an experiment is 1. (Page 205, Remark 2)
(v) The probability of an event is greater than or equal to 0 and less than or equal to 1. (Page 207)
2Which of the following experiments have equally likely outcomes? Explain.
Answer & Reference:
(i) A driver attempts to start a car. The car starts or does not start. → Not equally likely because the car may have mechanical issues.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot. → Not equally likely because skill affects outcome.
(iii) A trial is made to answer a true-false question. The answer is right or wrong. → Equally likely if guessing randomly.
(iv) A baby is born. It is a boy or a girl. → Approximately equally likely in general population. (Page 203)
3Why 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 & Reference:
Tossing a coin is fair because it has two equally likely outcomes - head and tail. Since the coin is unbiased, both teams have equal probability (1/2) of winning the toss. (Page 202)
4Which of the following cannot be the probability of an event? (A) 2/3 (B) -1.5 (C) 15% (D) 0.7
Answer & Reference: (B) -1.5 cannot be the probability of an event because probability always lies between 0 and 1 (inclusive). Negative values are not possible. (Page 207)
5If P(E) = 0.05, what is the probability of 'not E'?
6A 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?
7It 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?
8A 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?
9A 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?
10A 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?
11Gopi 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 & Reference:
Total fish = 5 + 8 = 13
P(male fish) = 5/13 (Page 215, Exercise 14.1)
12A 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?
13A 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.
Answer & Reference:
Possible outcomes: 1,2,3,4,5,6
(i) Prime numbers: 2,3,5 → P = 3/6 = 1/2
(ii) Numbers between 2 and 6: 3,4,5 → P = 3/6 = 1/2
(iii) Odd numbers: 1,3,5 → P = 3/6 = 1/2 (Page 215, Exercise 14.1)
14One 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
Answer & Reference:
Total cards = 52
(i) Red kings: 2 (hearts and diamonds) → P = 2/52 = 1/26
(ii) Face cards: 12 (Kings, Queens, Jacks) → P = 12/52 = 3/13
(iii) Red face cards: 6 (3 in hearts, 3 in diamonds) → P = 6/52 = 3/26
(iv) Jack of hearts: 1 → P = 1/52
(v) Spades: 13 → P = 13/52 = 1/4
(vi) Queen of diamonds: 1 → P = 1/52 (Page 207, Example 4)
15Five 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?
Answer & Reference:
(i) Total cards = 5, Queen = 1 → P(queen) = 1/5
(ii) After removing queen, total cards = 4
(a) P(ace) = 1/4
(b) P(queen) = 0/4 = 0 (impossible event) (Page 215, Exercise 14.1)
1612 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.
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?
Answer & Reference:
(i) Total bulbs = 20, Defective = 4 → P(defective) = 4/20 = 1/5
(ii) After removing 1 good bulb, total bulbs = 19, Good bulbs = 15 → P(not defective) = 15/19 (Page 216, Exercise 14.1)
18A 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.
Answer & Reference:
Total discs = 90
(i) Two-digit numbers: 90 - 9 = 81 → P = 81/90 = 9/10
(ii) Perfect squares: 1,4,9,16,25,36,49,64,81 → 9 numbers → P = 9/90 = 1/10
(iii) Numbers divisible by 5: 5,10,15,...,90 → 18 numbers → P = 18/90 = 1/5 (Page 216, Exercise 14.1)
19A 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?
Answer & Reference:
Total faces = 6
(i) A appears on 2 faces → P(A) = 2/6 = 1/3
(ii) D appears on 1 face → P(D) = 1/6 (Page 216, Exercise 14.1)
20Suppose 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?
Answer & Reference:
Fig. 14.6: Rectangular region with circle
Area of rectangle = 3 × 2 = 6 m²
Area of circle = πr² = π(0.5)² = π/4 m²
P(landing inside circle) = (π/4) / 6 = π/24 (Page 216, Exercise 14.1)
21A 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?
22(i) Complete the following table: [Probability table for sum on 2 dice] (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 & Reference:
(i) Completed table:
Event: 'Sum on 2 dice'
2
3
4
5
6
7
8
9
10
11
12
Probability
1/36
2/36
3/36
4/36
5/36
6/36
5/36
4/36
3/36
2/36
1/36
(ii) No, the argument is incorrect because the outcomes are not equally likely. For example, sum 7 can occur in 6 ways while sum 2 can occur in only 1 way. (Page 213, Example 13)
23A 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.
24A 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?
Answer & Reference:
Total outcomes = 6 × 6 = 36
(i) Outcomes without 5: 5 × 5 = 25 → P = 25/36
(ii) P(at least one 5) = 1 - P(no 5) = 1 - 25/36 = 11/36 (Page 216, Exercise 14.1)
25Which 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.
Answer & Reference:
(i) Incorrect. The outcomes are not equally likely. Actual outcomes: HH, HT, TH, TT. P(HH) = 1/4, P(TT) = 1/4, P(one of each) = 2/4 = 1/2.
(ii) Correct. Odd numbers: 1,3,5; Even numbers: 2,4,6. Both have 3 outcomes, so P(odd) = 3/6 = 1/2. (Page 217, Exercise 14.1)