Mathematics - Probability Question with Solution | TestHub

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. $$\text{ Then, }\mathrm{P}(\mathrm{X}=0)={}^4{\mathrm{C}}_0{\mathrm{p}}^0{\mathrm{q}}^4=\frac{16}{81}$$ $\Rightarrow {\mathrm{q}}^4={\left(\frac{2}{3}\right)}^4$ $\left(\because {}^n{C}_0=1\right)$ $\Rightarrow \mathrm{q}=\frac{2}{3}\Rightarrow \mathrm{p}=\frac{1}{3}$ $\therefore \mathrm{P}(\mathrm{X}=4)={}^4{\mathrm{C}}_4{\mathrm{p}}^4{\mathrm{q}}^0={\mathrm{p}}^4={\left(\frac{1}{3}\right)}^4=\frac{1}{81}$