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A multiple-choice exam has 5 questions, each with 3 options. Probability that a student gets at least 4 correct by random guessing:
A$10/243$
B$11/243$
C$5(1/3)^4(2/3) + (1/3)^5 = 11/243$
DBoth B and C
Answer & Solution
Correct answer: D. Both B and C
$p = 1/3$. $P(X \geq 4) = P(4) + P(5) = {}^5C_4(1/3)^4(2/3) + (1/3)^5 = 5 \cdot 2/243 + 1/243 = 11/243$.
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