A group consists of 12 persons, of which 3 are extremely patient, other 6 are extremely honest and rest are extremely kind. A person the group is selected at random. Assuming that each persons 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 values did you prefer more ?
The total number of persons = 12.
The number of persons who are extremely patient = 3.
The number of persons who are extremely kind = 12 – 3 – 6 = 3.
(i) P(selecting a person who is extremely patient) = \(\frac{number\, of \,favorable\,outcomes}{number\,of \,all\,possible\,outcomes}\) = \(\frac{3}{12}\) = \(\frac{1}{4}\)
Thus, the probability of selecting a person who is extremely patient is \(\frac{1}{4}\).
(ii) P(selecting a person who is extremely kind of honest) = \(\frac{number\, of \,favorable\,outcomes}{number\,of \,all\,possible\,outcomes}\) = \(\frac{6+3}{12}\) = \(\frac{9}{12}\)= \(\frac{3}{4}\)
Thus, the probability of selecting persons who is extremely kind or honest is \(\frac{3}{4}\).
from the three given values, we prefer honesty more.