A multiple-choice examination consists of 5 q | vedclass

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(C) Step $- 1$: Identify the parameters for the binomial distribution.Each question has $3$ choices with $1$ correct answer.Probability of success $p

Vedclass · Accepted Answer

$\frac{11}{3^5}$

Vedclass · Answer

(C) Step $- 1$: Identify the parameters for the binomial distribution.Each question has $3$ choices with $1$ correct answer.Probability of success $p = \frac{1}{3}$.Probability of failure $q = 1 - p = \frac{2}{3}$.Number of trials $n = 5$.Step $- 2$: Calculate the probability for $4$ or more correct answers.We need to find $P(X \ge 4) = P(X = 4) + P(X = 5)$.Using the binomial formula $P(X = k) = ^nC_k \cdot p^k \cdot q^{n-k}$:For $k = 4$:$P(X = 4) = ^5C_4 \cdot (\frac{1}{3})^4 \cdot (\frac{2}{3})^1 = 5 \cdot \frac{1}{81} \cdot \frac{2}{3} = \frac{10}{3^5}$.For $k = 5$:$P(X = 5) = ^5C_5 \cdot (\frac{1}{3})^5 \cdot (\frac{2}{3})^0 = 1 \cdot \frac{1}{3^5} \cdot 1 = \frac{1}{3^5}$.Step $- 3$: Sum the probabilities.$P(X \ge 4) = \frac{10}{3^5} + \frac{1}{3^5} = \frac{11}{3^5}$.Thus,the probability is $\frac{11}{3^5}$.

$A$ multiple-choice examination consists of $5$ questions. Each question has $3$ alternative answers,of which only $1$ is correct. What is the probability that a student will get $4$ or more correct answers?

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