Exercise 14.1
1Complete the following statements:
(i) Probability of an event E + Probability of the event ‘not E’ = ___ .
1
(ii) The probability of an event that cannot happen is ___ . Such an event is called ___ .
0; impossible event
(iii) The probability of an event that is certain to happen is ___ . Such an event is called ___ .
1; sure event (or certain event)
(iv) The sum of the probabilities of all the elementary events of an experiment is ___ .
1
(v) The probability of an event is greater than or equal to ___ and less than or equal to ___ .
0 and 1
2Which of the following experiments have equally likely outcomes? Explain.
(i) A driver attempts to start a car. The car starts or does not start.
Not equally likely
The outcome depends on factors like the condition of the car and fuel.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
Not equally likely
The outcome depends on the skill of the player.
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
Equally likely
There are only two possible answers, and each has the same chance.
(iv) A baby is born. It is a boy or a girl.
Equally likely
A baby has the same chance of being a boy or a girl.
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?
When a coin is tossed, there are only two outcomes, head and tail. Both are equally likely.
The result of a single toss cannot be predicted. So, each team has an equal chance of winning the toss.
So, tossing a coin is a fair way of deciding.
4Which of the following cannot be the probability of an event? (A) 23 (B) –1.5 (C) 15% (D) 0.7
The probability of an event always lies between 0 and 1 (0 ≤ P(E) ≤ 1). It can never be negative.
(B) –1.5
5If P(E) = 0.05, what is the probability of ‘not E’?
P(not E) = 1 – P(E) = 1 – 0.05 = 0.95
P(not E) = 0.95
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?
(i) an orange flavoured candy
There is no orange candy in the bag. So, it is an impossible event.
P(orange candy) = 0
(ii) a lemon flavoured candy
All candies are lemon flavoured. So, it is a sure event.
P(lemon candy) = 1
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?
P(same birthday) = 1 – P(not same birthday) = 1 – 0.992 = 0.008
Required probability 0.008
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?
Total number of balls = 3 + 5 = 8
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?
Total number of marbles = 5 + 8 + 4 = 17
(ii) white
P(white) = 817
(iii) not green
P(not green) = 1 – P(green) = 1 – 417 = 1317
1317
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 50 p coin? (ii) will not be a ₹ 5 coin?
Total number of coins = 100 + 50 + 20 + 10 = 180
(i) will be a 50 p coin
P(50 p coin) = 100180 = 59
59
(ii) will not be a ₹ 5 coin
P(not a ₹ 5 coin) = 1 – 10180 = 170180 = 1718
1718
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 (see Fig. 14.4). What is the probability that the fish taken out is a male fish?
Total number of fish = 5 + 8 = 13
P(male fish) = 513
513
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 (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?
Total number of possible outcomes = 8.
(ii) an odd number
Odd numbers are 1, 3, 5, 7, i.e., 4 outcomes.
12
(iii) a number greater than 2
Numbers greater than 2 are 3, 4, 5, 6, 7, 8, i.e., 6 outcomes.
34
(iv) a number less than 9
All 8 numbers are less than 9.
1
Advertisement
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.
Possible outcomes: 1, 2, 3, 4, 5, 6. Total = 6.
(i) a prime number
Prime numbers are 2, 3, 5.
P(prime number) = 36 = 12
12
(ii) a number lying between 2 and 6
Numbers between 2 and 6 are 3, 4, 5.
12
(iii) an odd number
Odd numbers are 1, 3, 5.
12
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
Total number of possible outcomes = 52.
(i) a king of red colour
There are 2 red kings (hearts and diamonds).
126
(ii) a face card
There are 12 face cards (4 kings, 4 queens, 4 jacks).
313
(iii) a red face card
There are 6 red face cards.
326
(iv) the jack of hearts
P = 152
(v) a spade
There are 13 spades.
14
(vi) the queen of diamonds
P = 152
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?
(ii) after the queen is put aside, 4 cards are left
(b) a queen
There is no queen left. So, it is an impossible event.
P(queen) = 0
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.
Total number of pens = 12 + 132 = 144
P(good pen) = 132144 = 1112
1112
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?
(ii)
One non-defective bulb is removed. Now there are 19 bulbs, of which 15 are not defective.
1519
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.
Total number of possible outcomes = 90.
(i) a two-digit number
Two-digit numbers are 10 to 90, i.e., 81 numbers.
910
(ii) a perfect square number
Perfect squares are 1, 4, 9, 16, 25, 36, 49, 64, 81, i.e., 9 numbers.
110
(iii) a number divisible by 5
Numbers divisible by 5 are 5, 10, ..., 90, i.e., 18 numbers.
15
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?
Total number of possible outcomes = 6.
(i) A
Two faces show A.
13
(ii) D
One face shows D.
P(D) = 16
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? (The rectangle measures 3 m × 2 m.)
Area of the rectangle = 3 × 2 = 6 m2
Area of the circle = πr2 = π × (12)2 = π4 m2
P = Area of circleArea of rectangle = π/46 = π24
π24
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?
Total pens = 144 ; good pens = 144 – 20 = 124
(i) She will buy it
P(good pen) = 124144 = 3136
3136
(ii) She will not buy it
P(defective pen) = 1 – 3136 = 536
536
22Refer to Example 13. (i) Complete the following table: (Event: ‘Sum on 2 dice’ 2, 3, 4, ..., 12; Probability: 136, ..., 536 (for 8), ..., 136) (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 111. Do you agree with this argument? Justify your answer.
When two dice are thrown together, there are 6 × 6 = 36 equally likely outcomes.
(i)
| Sum on 2 dice | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|
| Probability | 136 | 236 | 336 | 436 | 536 | 636 | 536 | 436 | 336 | 236 | 136 |
(ii)
No, I do not agree.
The 11 sums are not equally likely. For example, sum 2 comes only from (1, 1), but sum 7 comes from 6 outcomes. So, P(2) = 136 and P(7) = 636 = 16.
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.
Possible outcomes: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT. Total = 8.
Hanif wins in 2 outcomes (HHH and TTT).
P(win) = 28 = 14
P(lose) = 1 – 14 = 34
P(Hanif will lose) = 34
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?
Total number of outcomes = 6 × 6 = 36.
Outcomes in which 5 comes up 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). These are 11 outcomes.
(i) 5 will not come up either time
2536
(ii) 5 will come up at least once
1136
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 13.
Not correct
The possible outcomes are HH, HT, TH and TT. ‘One of each’ can happen in 2 ways. So, P(two heads) = 14, P(two tails) = 14 and P(one of each) = 24 = 12.
(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 12.
Correct
There are 3 odd numbers (1, 3, 5) and 3 even numbers (2, 4, 6) out of 6 outcomes. So, P(odd number) = 36 = 12.
