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(D) Let $X$ be the number of correct answers. The $3$ questions have probability of success $p_1 = \frac{1}{4}$ and failure $q_1 = \frac{3}{4}$. The $

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$\frac{3}{64}$

Vedclass · Answer

(D) Let $X$ be the number of correct answers. The $3$ questions have probability of success $p_1 = \frac{1}{4}$ and failure $q_1 = \frac{3}{4}$. The $2$ questions have probability of success $p_2 = \frac{1}{2}$ and failure $q_2 = \frac{1}{2}$.We want the probability of at least $4$ correct answers, i.e.,$P(X=4) + P(X=5)$.For $X=5$: All $5$ are correct. $P(X=5) = (\frac{1}{4})^3 \cdot (\frac{1}{2})^2 = \frac{1}{64} \cdot \frac{1}{4} = \frac{1}{256}$.For $X=4$: Either one of the $3$ questions is wrong, or one of the $2$ questions is wrong.Case $1$: One of the $3$ questions is wrong. Probability $= {^3C_2} \cdot (\frac{1}{4})^2 \cdot (\frac{3}{4})^1 \cdot (\frac{1}{2})^2 = 3 \cdot \frac{1}{16} \cdot \frac{3}{4} \cdot \frac{1}{4} = \frac{9}{256}$.Case $2$: One of the $2$ questions is wrong. Probability $= {^3C_3} \cdot (\frac{1}{4})^3 \cdot {^2C_1} \cdot (\frac{1}{2})^1 \cdot (\frac{1}{2})^1 = 1 \cdot \frac{1}{64} \cdot 2 \cdot \frac{1}{4} = \frac{2}{256}$.Total probability $= P(X=5) + P(X=4) = \frac{1}{256} + \frac{9}{256} + \frac{2}{256} = \frac{12}{256} = \frac{3}{64}$.

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