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A group consists of 12 persons, of which 3 are extremely patient, other 6 are extremely honest and rest are extremely kind. A person from the group is selected at random. Assuming that each person is equally likely to be selected, find the probability of selecting a person who is (i) extremely patient (ii) extremely kind or honest. Which of the above you prefer more.
Given:
A group consists of 12 persons, of which 3 are extremely patient, other 6 are extremely honest and rest are extremely kind. A person from the group is selected at random.
Each person is equally likely to be selected.
To do:
We have to find the probability of selecting a person who is (i) extremely patient (ii) extremely kind or honest.
Solution:
The total number of people in the group $n=12$.
Number of people who are extremely patient $=3$
Number of people who are extremely honest $=6$
This implies,
Number of people who are extremely kind $=12-(3+6)=12-9=3$
We know that,
Probability of an event $=\frac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ possible\ outcomes}$
(i) Number of people who are extremely patient $=3$
Total number of favourable outcomes $=3$.
Therefore,
The probability of selecting a person who is extremely patient $=\frac{3}{12}$
$=\frac{1}{4}$
(ii) Number of people who are extremely kind or honest $=3+6=9$
Total number of favourable outcomes $=9$.
Therefore,
The probability of selecting a person who is extremely kind or honest $=\frac{9}{12}$
$=\frac{3}{4}$
We prefer selecting a person who is extremely kind or honest.