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Question
If the sum of the mean and variance of a binomial distribution for 6 trials is \[\frac{10}{3},\] find the distribution.
Solution
Given that n = 6
The sum of mean and variance of a binomial distribution for 6 trials is \[\frac{10}{3}\]
\[\Rightarrow 6p + 6pq = \frac{10}{3}\]
\[ \Rightarrow 18p + 18p(1 - p) = 10\]
\[ \Rightarrow 18 p^2 - 36p + 10 = 0\]
\[ \Rightarrow (3p - 1)(6p - 10) = 0 \]
\[ \Rightarrow p = \frac{1}{3}\text{ or } \frac{5}{3}\]
\[p = \frac{5}{3} (\text{ Neglected as it is greater than 1} )\]
\[ \therefore p = \frac{1}{3}\]
\[ \Rightarrow q = 1 - p = \frac{2}{3}\]
\[\text{ Hence, the distribution is given by } \]
\[P(X = r) =^{6}{}{C}_r \left( \frac{1}{3} \right)^r \left( \frac{2}{3} \right)^{6 - r} , r = 0, 1, 2 . . . . . 6\]
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