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If in a binomial distribution $\mathrm{n}=4, \mathrm{P}(\mathrm{X}=0)=\frac{16}{81}$, then $\mathrm{P}(\mathrm{X}=4)$ equals
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The correct answer is:
$\frac{1}{81}$
Given $n=4$ and $P(X=0)=\frac{16}{81}$
Let $\mathrm{p}$ be the probability of success and $\mathrm{q}$ that of failure in a trial.
$$
\text { Then, } \mathrm{P}(\mathrm{X}=0)={ }^{4} \mathrm{C}_{0} \mathrm{p}^{0} \mathrm{q}^{4}=\frac{16}{81}
$$
$\Rightarrow \quad \mathrm{q}^{4}=\left(\frac{2}{3}\right)^{4}$
$\left(\because{ }^{n} C_{0}=1\right)$
$\Rightarrow \quad \mathrm{q}=\frac{2}{3} \Rightarrow \mathrm{p}=\frac{1}{3}$
$\therefore \quad \mathrm{P}(\mathrm{X}=4)={ }^{4} \mathrm{C}_{4} \mathrm{p}^{4} \mathrm{q}^{0}=\mathrm{p}^{4}=\left(\frac{1}{3}\right)^{4}=\frac{1}{81}$
Let $\mathrm{p}$ be the probability of success and $\mathrm{q}$ that of failure in a trial.
$$
\text { Then, } \mathrm{P}(\mathrm{X}=0)={ }^{4} \mathrm{C}_{0} \mathrm{p}^{0} \mathrm{q}^{4}=\frac{16}{81}
$$
$\Rightarrow \quad \mathrm{q}^{4}=\left(\frac{2}{3}\right)^{4}$
$\left(\because{ }^{n} C_{0}=1\right)$
$\Rightarrow \quad \mathrm{q}=\frac{2}{3} \Rightarrow \mathrm{p}=\frac{1}{3}$
$\therefore \quad \mathrm{P}(\mathrm{X}=4)={ }^{4} \mathrm{C}_{4} \mathrm{p}^{4} \mathrm{q}^{0}=\mathrm{p}^{4}=\left(\frac{1}{3}\right)^{4}=\frac{1}{81}$
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