In a company there are two vacancies. A man and his wife come for interview. The probability of selection of man is 1/10 and his wife is 1/8.Find the probability that only one of them be selected?
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Step-by-step explanation:
A man and his wife appear in an interview for two vacancies in the same post. The probability of husband’s selection is (1/7) and the probability of wife’s selection is (1/5). What is the assurance that only one of them is selected?
Let A = the event that the wife is selected
Let B= the event that the husband is selected
We have,
P(A)= 1/5
P(B)= 1/7
The event that only one of them is selected (let it be ‘C’) can be interpreted as;
Step-by-step explanation:
A man and his wife appear in an interview for two vacancies in the same post. The probability of husband’s selection is (1/7) and the probability of wife’s selection is (1/5). What is the assurance that only one of them is selected?
Let A = the event that the wife is selected
Let B= the event that the husband is selected
We have,
P(A)= 1/5
P(B)= 1/7
The event that only one of them is selected (let it be ‘C’) can be interpreted as;
The wife is selected and the husband is not. OR
The husband is selected and the wife is not.
Mathematically, statement 1 is equivalent to
P(A & not B) and statement 2 is equivalent to
P(B & not A).
Then , P(C) = P(A & not B) + P(B & not A)
= P(A)P(not B) + P(B) P(not A)……..(1)
since the events are independent.
P(not B) = 1- P(B) = 1–1/7= 6/7
P(not A) = 1- P(A) = 1–1/5= 4/5
.: Eqn 1 becomes:
P(C) = 1/5 × 6/7 + 1/7 × 4/5 = 0.284