CBSE Class 10 Mathematics Probability Assignment Set 05

Question. A bag contains 3 red balls and 5 black balls. A ball is drawn at random frm the bag. What is the probability that the ball drawn is (i) red? (ii) not red? (CBSE 2012)
Answer: Total number of balls = 3 + 5 = 8
∴ Number of all possible outcomes = 8
(i) For red balls:
∴ There are 3 red balls.
∴ Number of favourable outcomes = 3
∴ P_Red = Number of favourable outcomes/Number of all possible outcomes
= 3/8
(ii) For not red balls:
Probability of the ball drawn which is not red
= 1 - (3/8) = 8 - 3/8 = 5/8

Question. A piggy bank contains hundred 50p coins, fifty Re 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? and (ii) will not be ₹ 5 coin?
Answer: Number of:
50 p coins = 100
Re 1 coins = 50
₹ 2 coins = 20
₹ 5 coins = 10
Total number of coins = 100 + 50 + 20 + 5 = 180
(i) For a 50 p coin:
Favourable events = 100
∴ P_{(50 p)} = 100/180 = 5/9
(ii) For not a ₹ 5 coin:
ä Number of ₹ 5 coins = 10
∴ Number of ‘not ₹ 5’ coins = 180 − 10 = 170
⇒ Favourable outcomes = 170
∴ P(not 5 rupee coin) = 170/180 = 17/18

Question. 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: We have
Number of good pens = 132
Number of defective pens = 12
∴ Total number of pens = 132 + 12 = 144
For good pens:
ä There are 132 good pens
∴ Number of favourable outcomes = 132
⇒ P_{(good pens)} = Number of favourable outcomes/Total possible outcomes
= 132/144 = 11/12

Question. (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: Since, there are 20 bulbs in the lot.
∴ Total number of possible outcomes = 20
(i) ⸪ Number defective bulbs = 4
i.e., Favourable outcomes = 4
⇒ P_{(defective bulb)} = Number of favourable outcomes/Total number of outcomes
= 4/20 = 1/5
(ii) ⸪ The bulb drawn above is not included in the lot.
∴ Remaining number of bulbs = 20 − 1 = 19.
⇒ Total number of possible outcomes = 19.
⸪ Number of bulbs which are not defective = 19 − 4 = 15
⇒ Favourable number of outcomes = 15
∴ P_{(not defective bulb)} = Number of favourable outcomes/Total number of possible outcomes
= 15/19

Question. 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?
Answer: Total number of ball pens = 144
⇒ All possible outcomes = 144
(i) Since there are 20 defective pens
∴ Number of good pens 144 − 20 = 124
⇒ Number of favourable outcomes = 124
∴ Probability that she will buy it
= 124/144 = 31/36
(ii) Probability that she will not buy it
= 1 − [Probability that she will buy it]
= 1 − (31/36)
= 36 − 31 / 36 = 5/36

Question. 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: Let T denotes the tail and H denotes the head.
∴ All the possible outcomes are:
H H H,     H H T,     H T T,     T T T,     T T H,     T H T,     T T H,     H T H
∴ Number of all possible outcomes = 8
Let the event that Hanif will lose the game be denoted by E.
∴ Favourable events are:
H H T,     H T H,     T H H,     T H T ,     T T H,     H T T
⇒ Number of favourable outcomes = 3
∴ P (E) = 6/8 = 3/4

Question. 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?
[Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment.]
Answer: Since, throwing a die twice or throwing two dice simultaneously is the same.
∴ All possible outcomes are:
(1, 1); (1, 2); (1, 3); (1, 4); (1, 5); (1, 6)
(2, 1); (2, 2); (2, 3); (2, 4); (2, 5); (2, 6)
(3, 1); (3, 2); (3, 3); (3, 4); (3, 5); (3, 6)
(4, 1); (4, 2); (4, 3); (4, 4); (4, 5); (4, 6)
(5, 1); (5, 2); (5, 3); (5, 4); (5, 5); (5, 6)
(6, 1); (6, 2); (6, 3); (6, 4); (6, 5); (6, 6)
∴ All possible outcomes = 36
(i) Let E be the event that 5 does not come up either time, then
The favourable outcomes are [36 − (5 + 6)] = 25
⇒ P(E) = 25/36
(ii) Let N be the event that 5 will come up at least once, then Number of favourable outcomes = 5 + 6 = 11
∴ P(N) = 11/36

Question. 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.
Answer: (i) Not correct.
Because, the situation ‘one of each’ can result in two ways HT and TH.
∴ The probability = 1/4.
(ii) Correct.
Because the two outcomes are possible.