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In a binomial distribution, the mean is $\frac{2}{3}$ and the variance
is $\frac{5}{9} .$ What is the probability that $\mathrm{X}=2$ ?
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is $\frac{5}{9} .$ What is the probability that $\mathrm{X}=2$ ?
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The correct answer is:
$\frac{25}{216}$
Mean $=\frac{2}{3}$, variance $=\frac{5}{9}$
$n p=\frac{2}{3}, n p q=\frac{5}{9}$
$\Rightarrow \frac{2}{3} q=\frac{5}{9} \Rightarrow q=\frac{5}{9} \times \frac{3}{2}=\frac{5}{6}$
So, $\mathrm{p}=1-\frac{5}{6}=\frac{1}{6}$
Now, $\mathrm{np}=\frac{2}{3} \Rightarrow \mathrm{n}\left(\frac{1}{6}\right)=\frac{2}{3} \Rightarrow \mathrm{n}=\frac{2}{3} \times 6=4$
$\therefore \mathrm{p}(\mathrm{x}=2)={ }^{\mathrm{n}} \mathrm{c}_{\mathrm{r}} \cdot \mathrm{p}^{\mathrm{r}} \cdot \mathrm{q}^{\mathrm{n}-\mathrm{r}}={ }^{4} \mathrm{c}_{2} \cdot\left(\frac{1}{6}\right)^{2} \cdot\left(\frac{5}{6}\right)^{4-2}$
$=6 \times \frac{1}{36} \times \frac{25}{36}$
$=\frac{25}{216}$.
$n p=\frac{2}{3}, n p q=\frac{5}{9}$
$\Rightarrow \frac{2}{3} q=\frac{5}{9} \Rightarrow q=\frac{5}{9} \times \frac{3}{2}=\frac{5}{6}$
So, $\mathrm{p}=1-\frac{5}{6}=\frac{1}{6}$
Now, $\mathrm{np}=\frac{2}{3} \Rightarrow \mathrm{n}\left(\frac{1}{6}\right)=\frac{2}{3} \Rightarrow \mathrm{n}=\frac{2}{3} \times 6=4$
$\therefore \mathrm{p}(\mathrm{x}=2)={ }^{\mathrm{n}} \mathrm{c}_{\mathrm{r}} \cdot \mathrm{p}^{\mathrm{r}} \cdot \mathrm{q}^{\mathrm{n}-\mathrm{r}}={ }^{4} \mathrm{c}_{2} \cdot\left(\frac{1}{6}\right)^{2} \cdot\left(\frac{5}{6}\right)^{4-2}$
$=6 \times \frac{1}{36} \times \frac{25}{36}$
$=\frac{25}{216}$.
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