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A die is thrown 6 times. 'Getting an odd number' is a success. The probability of **exactly 5 successes** is:
A$3/32$
B$6/64$
C$6 \cdot (1/2)^6 = 6/64 = 3/32$
DBoth A and C
Answer & Solution
Correct answer: D. Both A and C
$p = 1/2, q = 1/2, n = 6$. $P(X = 5) = {}^6C_5 (1/2)^5 (1/2)^1 = 6/64 = 3/32$.
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