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If $X \sim B(4, p)$ and $P(X=0)=\frac{16}{81}$, then $P(X=4)=$
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Verified Answer
The correct answer is:
$\frac{1}{81}$
We have $\mathrm{n}=4$ and $\mathrm{P}(\mathrm{x}=0)=\frac{16}{81}$
$$
\begin{aligned}
& \therefore \frac{16}{81}={ }^4 \mathrm{C}_0(\mathrm{p})^4(\mathrm{q})^0 \\
& \therefore \frac{16}{81}=(\mathrm{p})^4 \Rightarrow \mathrm{P}=\frac{2}{3} \Rightarrow \mathrm{q}=\frac{1}{3} \\
& \therefore \mathrm{P}(\mathrm{x}=4)={ }^4 \mathrm{C}_4(\mathrm{p})^0(\mathrm{q})^4=(1)(1)\left(\frac{1}{3}\right)^4=\frac{1}{81}
\end{aligned}
$$
$$
\begin{aligned}
& \therefore \frac{16}{81}={ }^4 \mathrm{C}_0(\mathrm{p})^4(\mathrm{q})^0 \\
& \therefore \frac{16}{81}=(\mathrm{p})^4 \Rightarrow \mathrm{P}=\frac{2}{3} \Rightarrow \mathrm{q}=\frac{1}{3} \\
& \therefore \mathrm{P}(\mathrm{x}=4)={ }^4 \mathrm{C}_4(\mathrm{p})^0(\mathrm{q})^4=(1)(1)\left(\frac{1}{3}\right)^4=\frac{1}{81}
\end{aligned}
$$
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