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A group consists of 12 persons, of which 3 are extremely patients, 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 $( 1)$.extremely patient, $( 2)$.extremely kind or honest. which of the above value you prefer more?
Given: A group of 12 persons, 3 of them are extremely patient, 6 are extremely honest and rast of them are extremely kind.
To do: To find the probability of selecting a person who is $( 1)$.extremely patient, $( 2)$.extremely kind and honest.
Solution:
The group consists of 12 persons
Total number of possible outcomes = 12
Let A is the event of selecting persons who are extremely patient.
Number of outcomes favorable to A is 3.
Let B is the event of selecting persons who are extremely kind or honest.
Number of persons who are extremely honest is 6.
Number of persons who are extremely kind is 12 - (6+ 3) = 3 .
Number of outcomes favourable to $\displaystyle B\ =\ 6+3=9$.
$( 1) .$ $P( A) =\frac{Number\ of\ outcomes\ favorable\ to\ A}{total\ number\ of\ possible\ outcomes}$
$=\frac{3}{12}$
$=\frac{1}{4}$
$( 2) $ $P( A) =\frac{Number\ of\ outcomes\ favourable\ to\ B}{total\ number\ of\ possible\ outcomes}$
$=\frac{9}{12}$
$=\frac{3}{4}$
Each of the three values, patience, honesty and kindness is important in one's life
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