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(C) Let $X$ be the number of correct answers. Since each question has two options (True or False), the probability of a correct answer is $p = \frac{1

Vedclass · Accepted Answer

$\frac{11}{32}$

Vedclass · Answer

(C) Let $X$ be the number of correct answers. Since each question has two options (True or False), the probability of a correct answer is $p = \frac{1}{2}$ and the probability of an incorrect answer is $q = \frac{1}{2}$.This follows a binomial distribution $B(n, p)$ where $n = 6$ and $p = \frac{1}{2}$.The student passes if he gets $4, 5,$ or $6$ correct answers.The probability of passing is $P(X \ge 4) = P(X=4) + P(X=5) + P(X=6)$.Using the formula $P(X=k) = {}^nC_k p^k q^{n-k}$:$P(X=4) = {}^6C_4 (\frac{1}{2})^4 (\frac{1}{2})^2 = 15 	imes (\frac{1}{2})^6 = \frac{15}{64}$.$P(X=5) = {}^6C_5 (\frac{1}{2})^5 (\frac{1}{2})^1 = 6 	imes (\frac{1}{2})^6 = \frac{6}{64}$.$P(X=6) = {}^6C_6 (\frac{1}{2})^6 (\frac{1}{2})^0 = 1 	imes (\frac{1}{2})^6 = \frac{1}{64}$.Total probability $= \frac{15+6+1}{64} = \frac{22}{64} = \frac{11}{32}$.

$A$ student is given $6$ questions in an examination with true or false type of answers. If he writes $4$ or more correct answers, he passes in the examination. The probability that he passes in the examination is

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