Let X have a binomial distribution B(n, p) such that the sum and the product of the mean and variance of X are 24 and 128 respectively. If $P(X>n-3)=\frac{k}{2^{n}}$, then k is equal to :
Solution
<p>Mean $= np = 16$</p>
<p>Variance $= npq = 8$</p>
<p>$\Rightarrow q = p = {1 \over 2}$ and $n = 32$</p>
<p>$P(x > n - 3) = p(x = n - 2) + p(x = n - 1) + p(x = n)$</p>
<p>$$ = \left( {{}^{32}{C_2} + {}^{32}{C_1} + {}^{32}{C_0}} \right)\,.\,{1 \over {{2^n}}}$$</p>
<p>$= {{529} \over {{2^n}}}$</p>
About this question
Subject: Mathematics · Chapter: Probability · Topic: Random Variables and Distributions
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