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A random experiment is conducted five times. If the number of successes of the experiment follows binomial distribution such that the difference of mean and variance of the successes is $\frac{5}{9}$, then the probability of getting atmost two successes is
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The correct answer is:
$\frac{64}{81}$
We have,
$\begin{aligned} & n=5 \text { and } n p-n p q=\frac{5}{9} \\ & \therefore \quad n p-n p q=\frac{5}{9} \\ & \Rightarrow \quad 5 p-5 p q=\frac{5}{9} \Rightarrow p-p q=\frac{1}{9} \\ & \Rightarrow \quad p(1-q)=\frac{1}{9} \Rightarrow p^2=\frac{1}{9} \Rightarrow p=\frac{1}{3} \\ & \therefore \quad q=1-p=1-\frac{1}{3}=\frac{2}{3} \\ & \therefore \text { Required probability }=P(x \leq 2) \\ & \quad=P(x=0)+P(x=1)+P(x=2) \\ & \quad={ }^5 C_0\left(\frac{1}{3}\right)^0\left(\frac{2}{3}\right)^5+{ }^5 C_1\left(\frac{1}{3}\right)^1\left(\frac{2}{3}\right)^4+{ }^5 C_2\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)^3 \\ & \quad=\frac{32}{243}+\frac{80}{243}+\frac{80}{243}=\frac{192}{243}=\frac{64}{81}\end{aligned}$
$\begin{aligned} & n=5 \text { and } n p-n p q=\frac{5}{9} \\ & \therefore \quad n p-n p q=\frac{5}{9} \\ & \Rightarrow \quad 5 p-5 p q=\frac{5}{9} \Rightarrow p-p q=\frac{1}{9} \\ & \Rightarrow \quad p(1-q)=\frac{1}{9} \Rightarrow p^2=\frac{1}{9} \Rightarrow p=\frac{1}{3} \\ & \therefore \quad q=1-p=1-\frac{1}{3}=\frac{2}{3} \\ & \therefore \text { Required probability }=P(x \leq 2) \\ & \quad=P(x=0)+P(x=1)+P(x=2) \\ & \quad={ }^5 C_0\left(\frac{1}{3}\right)^0\left(\frac{2}{3}\right)^5+{ }^5 C_1\left(\frac{1}{3}\right)^1\left(\frac{2}{3}\right)^4+{ }^5 C_2\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)^3 \\ & \quad=\frac{32}{243}+\frac{80}{243}+\frac{80}{243}=\frac{192}{243}=\frac{64}{81}\end{aligned}$
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