If $X \sim B(4, p)$ and $P(X=0)=\frac{16}{81}$, then $P(X=4)=$

If $X \sim B(4, p)$ and $P(X=0)=\frac{16}{81}$, then $P(X=4)=$
  1. $\frac{1}{81}$
  2. $\frac{1}{16}$
  3. $\frac{1}{8}$
  4. $\frac{1}{27}$

Solution

We have $\mathrm{n}=4$ and $\mathrm{P}(\mathrm{x}=0)=\frac{16}{81}$ $\begin{aligned} & \therefore \frac{16}{81}={ }^4 \mathrm{C}_0(\mathrm{p})^4(\mathrm{q})^0 \\ & \therefore \frac{16}{81}=(\mathrm{p})^4 \Rightarrow \mathrm{P}=\frac{2}{3} \Rightarrow \mathrm{q}=\frac{1}{3} \\ & \therefore \mathrm{P}(\mathrm{x}=4)={ }^4 \mathrm{C}_4(\mathrm{p})^0(\mathrm{q})^4=(1)(1)\left(\frac{1}{3}\right)^4=\frac{1}{81} \end{aligned}$

Asked in: MHT CET 2021 (20 Sep Shift 1)

Practice more Probability questions on Aicharya