Question

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. Find the probability that the selected person is (i) extremely patient (ii) extremely kind or honest. Which of the above you will prefer more.

Answer 1

Total no. of people in the group $=12$ $(3$ extremely patient, $6$ extremely honest and $3$ extremely kind$)$

Solution(i):

No. of extremely patient people $=3$

Therefore, ${}^3{C}_1$( Selecting $1$ out of $3$ items) times out of ${}^{12}{C}_1$( Selecting $1$ out of $12$ items) an extremely patient person is picked.

i.e.,

Let $E_1$ be the event of selecting an extremely patient person

We know that, $P\left(E_1\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{{}^3{C}_1}{{}^{12}{C}_1}$$=\frac{3}{12}$$=\frac{1}{4}$

Solution(ii):

No. of extremely honest or kind people $=9$ $(6$ extremely honest & $3$ extremely kind$)$

Therefore, ${}^9{C}_1$( Selecting $1$ out of $3$ items) times out of ${}^{12}{C}_1$( Selecting $1$ out of $12$ items)an extremely honest or kind person is picked.

i.e.,

Let $E_2$ be the event of selecting an extremely honest or kind person

We know that, $P\left(E_2\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{{}^9{C}_1}{{}^{12}{C}_1}$$=\frac{9}{12}$$=\frac{3}{4}$

An extremely honest or kind person is preferred more as the probability of selection is more than that of an extremely patient person. ....... [$P\left(E_2\right)>P\left(E_1\right)$].