A group consists of 12 persons, of which 3 are extremely patient, other 6 are extremely honest and rest are extremely kind. A person form 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.
Answers
SOLUTION :
Given : Total number of person in a group = 12
Total number of possible outcomes = 12
Number of persons who are extremely patient = 3
Number of persons who are extremely honest = 6
Number of person who are extremely kind = 12 -(3+6) = 12 - 9 = 3
(i) Let E1 = Event of selecting a person who is extremely patient
Number of persons who are extremely patient = 3
Number of favorable outcomes = 3
Probability (E1) = Number of favourable outcomes / Total number of outcomes
P(E1) = 3/12 = 1/4
Hence, the required probability of selecting a person who is extremely patient ,P(E1) = 1/2 .
(ii) Let E2 = Event of selecting a person who are extremely kind or honest
Number of persons who are extremely kind or honest = 3 + 6 = 9
Number of favorable outcomes = 9
Probability (E2) = Number of favourable outcomes / Total number of outcomes
P(E2) = 9/12 = 3/4
Hence, the required probability of selecting a person who is extremely kind or honest ,P(E2) = 3/4.
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Answer: i) 1/2 ii) 3/4
Step-by-step explanation:
let n(total) be total number of people= 12
n(P) be number of people extremely patient= 3
n(H) be number of people extremely honest= 6
n(K) be number of people extremely kind = 12-[n(H)+n(P)]= 12-9= 3
i) P(extremely patient= n(P)/n(total) =3/12= 1/4
ii) P(extremely honest or kind)= n(H)+n(K)/n(total)= 9/12 3/4