An examination is attempted by 5000 graduates | vedclass

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Question

(A) Let $E_1$,$E_2$,and $E_3$ be the events that the selected examinee is a graduate,a post-graduate,and a doctorate holder,respectively.Total examine

Vedclass · Accepted Answer

$\frac{45}{169}$

Vedclass · Answer

(A) Let $E_1$,$E_2$,and $E_3$ be the events that the selected examinee is a graduate,a post-graduate,and a doctorate holder,respectively.Total examinees = $5000 + 2000 + 1000 = 8000$.$P(E_1) = \frac{5000}{8000} = \frac{5}{8}$,$P(E_2) = \frac{2000}{8000} = \frac{2}{8} = \frac{1}{4}$,$P(E_3) = \frac{1000}{8000} = \frac{1}{8}$.Let $A$ be the event that the examinee passed the examination.$P(A|E_1) = \frac{2}{3}$,$P(A|E_2) = \frac{3}{4}$,$P(A|E_3) = \frac{4}{5}$.Using Bayes' Theorem,the probability that the examinee is a post-graduate given that they passed is:$P(E_2|A) = \frac{P(E_2) 	imes P(A|E_2)}{P(E_1) 	imes P(A|E_1) + P(E_2) 	imes P(A|E_2) + P(E_3) 	imes P(A|E_3)}$$P(E_2|A) = \frac{\frac{2}{8} 	imes \frac{3}{4}}{\frac{5}{8} 	imes \frac{2}{3} + \frac{2}{8} 	imes \frac{3}{4} + \frac{1}{8} 	imes \frac{4}{5}}$$P(E_2|A) = \frac{\frac{6}{32}}{\frac{10}{24} + \frac{6}{32} + \frac{4}{40}} = \frac{\frac{3}{16}}{\frac{5}{12} + \frac{3}{16} + \frac{1}{10}}$Finding a common denominator $(240)$:$P(E_2|A) = \frac{\frac{45}{240}}{\frac{100}{240} + \frac{45}{240} + \frac{24}{240}} = \frac{45}{100 + 45 + 24} = \frac{45}{169}$.

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