The mean and variance of a binomial distribution are $\alpha$ and $\frac{\alpha}{3}$ respectively. If $\mathrm{P}(X=1)=\frac{4}{243}$, then $\mathrm{P}(X=4$ or 5$)$ is equal to :
Solution
<p>Given, mean $= np = \alpha$.</p>
<p>and variance $= npq = {\alpha \over 3}$</p>
<p>$\Rightarrow q = {1 \over 3}$ and $p = {2 \over 3}$</p>
<p>$P(X = 1) = n.{p^1}.{q^{n - 1}} = {4 \over {243}}$</p>
<p>$$ \Rightarrow n.{2 \over 3}.{\left( {{1 \over 3}} \right)^{n - 1}} = {4 \over {243}}$$</p>
<p>$\Rightarrow n = 6$</p>
<p>$$P(X = 4\,\mathrm{or}\,5) = {}^6{C_4}\,.\,{\left( {{2 \over 3}} \right)^4}\,.\,{\left( {{1 \over 3}} \right)^2} + {}^6{C_5}\,.\,{\left( {{2 \over 5}} \right)^5}\,.\,{1 \over 3}$$</p>
<p>$= {{16} \over {27}}$</p>
About this question
Subject: Mathematics · Chapter: Probability · Topic: Classical and Axiomatic Probability
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