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Question
If the mean and variance of a binomial distribution are 4 and 3, respectively, the probability of getting exactly six successes in this distribution is
Options
\[^{16}{}{C}_6 \left( \frac{1}{4} \right)^{10} \left( \frac{3}{4} \right)^6\]
\[^{16}{}{C}_6 \left( \frac{1}{4} \right)^6 \left( \frac{3}{4} \right)^{10}\]
\[^{12}{}{C}_6 \left( \frac{1}{20} \right) \left( \frac{3}{4} \right)^6\]
\[^{12}{}{C}_6 \left( \frac{1}{4} \right)^6 \left( \frac{3}{4} \right)^6\]
Solution
\[^{16}{}{C}_6 \left( \frac{1}{4} \right)^6 \left( \frac{3}{4} \right)^{10}\]
Mean (np) = 4 and Variance (npq) = 3
\[\therefore q = \frac{3}{4}\]
\[ \Rightarrow p = 1 - \frac{3}{4} = \frac{1}{4}\text{ and } n = 16\]
\[\text{ Let X denotes the number of successes in 16 trials . Then, } \]
\[P(X = r) = ^{16}{}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^{16 - r} \]
\[ \Rightarrow P(X = 6) = \text{ Probability (getting exactly 6 successes } )\]
\[ = 16 C_6 \left( \frac{1}{4} \right)^6 \left( \frac{3}{4} \right)^{10} \]
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