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Question
If in a binomial distribution n = 4, P (X = 0) = \[\frac{16}{81}\], then P (X = 4) equals
Options
\[\frac{1}{16}\]
\[\frac{1}{81}\]
\[\frac{1}{27}\]
\[\frac{1}{8}\]
Solution
\[\frac{1}{81}\] In the given binomial distribution, n = 4 and
\[P(X = 0) = \frac{16}{81} \]
\[\text{ Binomial distribution is given by } \]
\[P(X = 0) =^ {4}{}{C}_0 \ p^0 q^{4 - 0} = q^4 \]
\[\text{ We know that P } (X = 0) = \frac{16}{81} \]
\[ \therefore q^4 = \frac{16}{81}\]
\[ \Rightarrow q^4 = \left( \frac{2}{3} \right)^4 \]
\[ \Rightarrow q = \frac{2}{3}\]
\[ \therefore p = 1 - \frac{2}{3} = \frac{1}{3}\]
\[\text{ Then } , P(X = 4) = ^{4}{}{C}_4 \ p^4 q^{4 - 4} \]
\[ = \left( \frac{1}{3} \right)^4 \]
\[ = \frac{1}{81}\]
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