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If \(\mathrm{X}\) follows Binomial distribution with mean 3 and variance 2 , then \(P(X \geq 8)\) is equal to :
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The correct answer is:
\(\frac{19}{3^9}\)
As given,
\(\text {Mean }=n p=3 \text { and Variance }=n p q=2\)
\(\begin{aligned} & \therefore \quad \frac{\mathrm{npq}}{\mathrm{np}}=\frac{2}{3} \Rightarrow \mathrm{q}=\frac{2}{3} \\ & \text { and } \mathrm{p}=(1-\mathrm{q})=1-\frac{2}{3}-\frac{1}{3} \\ & \therefore \quad \mathrm{n} \cdot \frac{1}{3}=3 \Rightarrow \mathrm{n}=9 \\ & \therefore \quad \mathrm{P}(\mathrm{X} \geq 8)=\mathrm{P}(\mathrm{X}=8)+\mathrm{P}(\mathrm{X}=9) \\ & ={ }^9 \mathrm{C}_8\left(\frac{1}{3}\right)^8 \times \frac{2}{3}+{ }^9 \mathrm{C}_9\left(\frac{1}{3}\right)^9 \\ & =9\left(\frac{1}{3}\right)^8 \frac{2}{3}+1\left(\frac{1}{3}\right)^9=\frac{18+1}{3^9}=\frac{19}{3^9}\end{aligned}\)
\(\text {Mean }=n p=3 \text { and Variance }=n p q=2\)
\(\begin{aligned} & \therefore \quad \frac{\mathrm{npq}}{\mathrm{np}}=\frac{2}{3} \Rightarrow \mathrm{q}=\frac{2}{3} \\ & \text { and } \mathrm{p}=(1-\mathrm{q})=1-\frac{2}{3}-\frac{1}{3} \\ & \therefore \quad \mathrm{n} \cdot \frac{1}{3}=3 \Rightarrow \mathrm{n}=9 \\ & \therefore \quad \mathrm{P}(\mathrm{X} \geq 8)=\mathrm{P}(\mathrm{X}=8)+\mathrm{P}(\mathrm{X}=9) \\ & ={ }^9 \mathrm{C}_8\left(\frac{1}{3}\right)^8 \times \frac{2}{3}+{ }^9 \mathrm{C}_9\left(\frac{1}{3}\right)^9 \\ & =9\left(\frac{1}{3}\right)^8 \frac{2}{3}+1\left(\frac{1}{3}\right)^9=\frac{18+1}{3^9}=\frac{19}{3^9}\end{aligned}\)
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