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In a Binominal distribution, the mean is three times its variance. What is the probability of exactly 3 successes out of 5 trials? $\quad$
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Verified Answer
The correct answer is:
$\frac{80}{243}$
Mean $=\mathrm{np}$
Variance $=\mathrm{npq}$
Given, $n p=3 n p q$ $\Rightarrow q=\frac{1}{3}, p=\frac{2}{3}$.
Also, Given $n=5$ trials.
$r=3$
we know, $\mathrm{p}(\mathrm{x}=\mathrm{r})={ }^{\mathrm{n}} \mathrm{c}_{\mathrm{r}} \cdot \mathrm{p}^{\mathrm{r}} \cdot \mathrm{q}^{\mathrm{n}-\mathrm{r}}$
$\mathrm{p}(\mathrm{x}=3)={ }^{5} \mathrm{c}_{3}\left(\frac{2}{3}\right)^{3}\left(\frac{1}{3}\right)^{2}$
$=\frac{80}{243} .$
Variance $=\mathrm{npq}$
Given, $n p=3 n p q$ $\Rightarrow q=\frac{1}{3}, p=\frac{2}{3}$.
Also, Given $n=5$ trials.
$r=3$
we know, $\mathrm{p}(\mathrm{x}=\mathrm{r})={ }^{\mathrm{n}} \mathrm{c}_{\mathrm{r}} \cdot \mathrm{p}^{\mathrm{r}} \cdot \mathrm{q}^{\mathrm{n}-\mathrm{r}}$
$\mathrm{p}(\mathrm{x}=3)={ }^{5} \mathrm{c}_{3}\left(\frac{2}{3}\right)^{3}\left(\frac{1}{3}\right)^{2}$
$=\frac{80}{243} .$
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