Question:
Let \( X \) have the binomial distribution \( \mathrm{B}(n, p) \). If its mean is 3 and variance is 2, then \( \mathrm{P}\left(X < \frac{n}{4}\right) \) is equal to :
Let \( X \) have the binomial distribution \( \mathrm{B}(n, p) \). If its mean is 3 and variance is 2, then \( \mathrm{P}\left(X < \frac{n}{4}\right) \) is equal to :
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Use mean \( = np \) and variance \( = npq \) to find \( n \) and \( p \), then add the probabilities of the allowed values of \( X \).
Updated On: Nov 12, 2026
- \( \frac{29 \times 2^8}{3^9} \)
- \( \frac{25 \times 2^9}{3^9} \)
- \( \frac{163}{3^9} \)
- \( \frac{835}{3^9} \)
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The Correct Option is A
Solution and Explanation
Mean \( np = 3 \) and variance \( npq = 2 \), so \( q = \frac{2}{3},\ p=\frac{1}{3},\ n = 9 \).
\( \frac{n}{4} = 2.25 \), so \( X < 2.25 \) means \( X = 0, 1, 2 \).
\( P(X=0) = \left(\frac{2}{3}\right)^9 = \frac{2^9}{3^9} \)
\( P(X=1) = 9\cdot\frac{1}{3}\cdot\left(\frac{2}{3}\right)^8 = \frac{9\cdot 2^8}{3^9} \)
\( P(X=2) = 36\cdot\frac{1}{9}\cdot\left(\frac{2}{3}\right)^7 = \frac{36\cdot 2^7}{3^9} \)
Sum \( = \frac{512 + 2304 + 4608}{3^9} = \frac{7424}{3^9} = \frac{29\times 2^8}{3^9} \).
Answer: (1) \( \frac{29 \times 2^8}{3^9} \)
\( \frac{n}{4} = 2.25 \), so \( X < 2.25 \) means \( X = 0, 1, 2 \).
\( P(X=0) = \left(\frac{2}{3}\right)^9 = \frac{2^9}{3^9} \)
\( P(X=1) = 9\cdot\frac{1}{3}\cdot\left(\frac{2}{3}\right)^8 = \frac{9\cdot 2^8}{3^9} \)
\( P(X=2) = 36\cdot\frac{1}{9}\cdot\left(\frac{2}{3}\right)^7 = \frac{36\cdot 2^7}{3^9} \)
Sum \( = \frac{512 + 2304 + 4608}{3^9} = \frac{7424}{3^9} = \frac{29\times 2^8}{3^9} \).
Answer: (1) \( \frac{29 \times 2^8}{3^9} \)
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