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Question
For the vacancy advertised in the newspaper, 3000 candidates submitted their applications. From the data it was revealed that two third of the total applicants were females and other were males. The selection for the job was done through a written test. The performance of the applicants indicates that the probability of a male getting a distinction in written test is 0.4 and that a female gettinga distinction is 0.35. Find the probability that the candidate chosen at random will have a distinction in the written test.
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Solution
Given: Total candidates = 3000
No. of females = `2/3 xx 3000`
= 2000
No. of males = `1/3 xx 3000`
= 1000
E1: Chosen candidate is female
E2: Chosen candidate is male
A: Getting a distinction written test
∴ `P(E_1) = 2/3`,
`P(E_2) = 1/3`
P(A/E1) = 0.35
P(A/E2) = 0.4
From the total probability theorem
P(A) = P(E1) P(A/E1) + P(E2) P(A/E2)
= `2/3 xx 0.35 + 1/3 xx 0.4`
= `2/3 xx 35/100 + 1/3 xx 4/10`
= `(70/300 + 4/30)`
= `((70 + 40)/300)`
= `110/300`
= `11/30`
