Total number of pens = 144
Total number of defective pens = 20
Total number of good pens = 144 − 20 = 124
(i) Probability of getting a good pen
P (Nuri buys a pen)
(ii) P (Nuri will not buy a pen
(i) It can be observed that,
To get the sum as 2, possible outcomes = (1, 1)
To get the sum as 3, possible outcomes = (2, 1) and (1, 2)
To get the sum as 4, possible outcomes = (3, 1), (1, 3), (2, 2)
To get the sum as 5, possible outcomes = (4, 1), (1, 4), (2, 3), (3, 2)
To get the sum as 6, possible outcomes = (5, 1), (1, 5), (2, 4), (4, 2),
(3, 3)
To get the sum as 7, possible outcomes = (6, 1), (1, 6), (2, 5), (5, 2),
(3, 4), (4, 3)
To get the sum as 8, possible outcomes = (6, 2), (2, 6), (3, 5), (5, 3),
(4, 4)
To get the sum as 9, possible outcomes = (3, 6), (6, 3), (4, 5), (5, 4)
To get the sum as 10, possible outcomes = (4, 6), (6, 4), (5, 5)
To get the sum as 11, possible outcomes = (5, 6), (6, 5)
To get the sum as 12, possible outcomes = (6, 6)
Event:
Sum of two dice |
2
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3
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4
|
5
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6
|
7
|
8
|
9
|
10
|
11
|
12
|
Probability |
|
|
|
|
|
|
|
|
|
|
|
(ii)Probability of each of these sums will not be
as these sums are not equally likely.
Question 23:
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.
The possible outcomes are
{HHH, TTT, HHT, HTH, THH, TTH, THT, HTT}
Number of total possible outcomes = 8
Number of favourable outcomes = 2 {i.e., TTT and HHH}
P (Hanif will win the game)
P (Hanif will lose the game
Question 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?
[Hint: Throwinga die twice and throwing two dice simultaneously are treated as the same experiment].
Total number of outcomes = 6 × 6
= 36
(i)Total number of outcomes when 5 comes up on either time are (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (1, 5), (2, 5), (3, 5), (4, 5), (6, 5)
Hence, total number of favourable cases = 11
P (5 will come up either time)
P (5 will not come up either time)
(ii)Total number of cases, when 5 can come at least once = 11
P (5 will come at least once)
Question 25:
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.
(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.
(i) Incorrect
When two coins are tossed, the possible outcomes are (H, H), (H, T), (T, H), and (T, T). It can be observed that there can be one of each in two possible ways − (H, T), (T, H).
Therefore, the probability of getting two heads is
, the probability of getting two tails is
, and the probability of getting one of each is.
It can be observed that for each outcome, the probability is not.
(ii) Correct
When a dice is thrown, the possible outcomes are 1, 2, 3, 4, 5, and 6. Out of these, 1, 3, 5 are odd and 2, 4, 6 are even numbers.
Therefore, the probability of getting an odd number is.