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.

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Updated on: 10-Oct-2022

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