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(D) Let $S_1$ be the set of $4$ questions with probability of success $p_1 = \frac{3}{4}$ and $S_2$ be the set of $6$ questions with probability of su

Vedclass · Accepted Answer

$479$

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(D) Let $S_1$ be the set of $4$ questions with probability of success $p_1 = \frac{3}{4}$ and $S_2$ be the set of $6$ questions with probability of success $p_2 = \frac{1}{4}$.To get exactly $8$ questions correct,we consider the following cases $(x, y)$ where $x$ is the number of correct answers from $S_1$ and $y$ is the number of correct answers from $S_2$,such that $x+y=8$:Case $1$: $x=4, y=4$. Probability $= \binom{4}{4} (\frac{3}{4})^4 (\frac{1}{4})^0 	imes \binom{6}{4} (\frac{1}{4})^4 (\frac{3}{4})^2 = 1 	imes \frac{81}{256} 	imes 15 	imes \frac{9}{4096} = \frac{10935}{4^{10}}$.Case $2$: $x=3, y=5$. Probability $= \binom{4}{3} (\frac{3}{4})^3 (\frac{1}{4})^1 	imes \binom{6}{5} (\frac{1}{4})^5 (\frac{3}{4})^1 = 4 	imes \frac{27}{64} 	imes \frac{1}{4} 	imes 6 	imes \frac{3}{4096} = \frac{1944}{4^{10}}$.Case $3$: $x=2, y=6$. Probability $= \binom{4}{2} (\frac{3}{4})^2 (\frac{1}{4})^2 	imes \binom{6}{6} (\frac{1}{4})^6 (\frac{3}{4})^0 = 6 	imes \frac{9}{16} 	imes \frac{1}{16} 	imes 1 	imes \frac{1}{4096} = \frac{54}{4^{10}}$.Total probability $= \frac{10935 + 1944 + 54}{4^{10}} = \frac{12933}{4^{10}}$.Given $\frac{27k}{4^{10}} = \frac{12933}{4^{10}}$,we have $27k = 12933 \Rightarrow k = \frac{12933}{27} = 479$.

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