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(C) Let $E_1$ be the event that the student answers without guessing,and $E_2$ be the event that the student answers by guessing. Given $P(E_2) = 3/7$

Vedclass · Accepted Answer

$\frac{1024}{1429}$

Vedclass · Answer

(C) Let $E_1$ be the event that the student answers without guessing,and $E_2$ be the event that the student answers by guessing. Given $P(E_2) = 3/7$,so $P(E_1) = 1 - 3/7 = 4/7$.Let $A$ be the event that at least $3$ questions are answered correctly.For $E_1$ (without guessing),the probability of success is $p = 2/3$. Using binomial distribution $B(4, 2/3)$:$P(A|E_1) = \binom{4}{3} (2/3)^3 (1/3)^1 + \binom{4}{4} (2/3)^4 = 4 \cdot (8/27) \cdot (1/3) + 16/81 = 32/81 + 16/81 = 48/81 = 16/27$.For $E_2$ (guessing),the probability of success is $p = 1/2$. Using binomial distribution $B(4, 1/2)$:$P(A|E_2) = \binom{4}{3} (1/2)^4 + \binom{4}{4} (1/2)^4 = 5/16$.Using Bayes' Theorem,we need $P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}$.$P(E_1|A) = \frac{(4/7) \cdot (16/27)}{(4/7) \cdot (16/27) + (3/7) \cdot (5/16)} = \frac{64/189}{64/189 + 15/112} = \frac{64/189}{7168/21168 + 2835/21168} = \frac{64/189}{10003/21168} = \frac{64 \cdot 112}{10003} = \frac{7168}{10003} = \frac{1024}{1429}$.

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