Mathematics - Probability Question with Solution | TestHub

Given that $P\left(X_1\right)=\frac{1}{2},\mathrm{P}\left(X_2\right)=\frac{1}{4},\mathrm{P}\left(X_3\right)=\frac{1}{4}$ $P(X)=P$ (at least 2 engines are functioning) $\begin{array}{rl}=P(X_1\cap X_2& \cap X_3^C)+P\left(X_1\cap X_2^C\cap X_3\right)\\ & +P\left(X_1^C\cap X_2\cap X_3\right)+P\left(X_1\cap X_2\cap X_3\right)\end{array}$ $=\frac{1}{2}\times \frac{1}{4}\times \frac{3}{4}+\frac{1}{2}\times \frac{3}{4}\times \frac{1}{4}+\frac{1}{2}\times \frac{1}{4}\times \frac{1}{4}+\frac{1}{2}\times \frac{1}{4}\times \frac{1}{4}=\frac{1}{4}$ (a) $P\left(X_1^C/X\right)=\frac{P\left(X_1^C\cap X\right)}{P(X)}=\frac{P\left(X_1^C\cap X_2\cap X_3\right)}{P(X)}$ $=\frac{\frac{1}{2}\times \frac{1}{4}\times \frac{1}{4}}{\frac{1}{4}}=\frac{1}{8}$ $\therefore$ (a) is not true. (b) $P$ [Exactly two engines are functioning $/X]$ $=\frac{P[(\text{ Exactly two engines are functioning })\cap X]}{P(X)}$ $=\frac{P\left(X_1^C\cap X_2\cap X_3\right)+P\left(X_1\cap X_2^C\cap X_3\right)+P\left(X_1\cap X_2\cap X_3^C\right)}{P(X)}$ $=\frac{\frac{1}{2}\times \frac{1}{4}\times \frac{1}{4}+\frac{1}{2}\times \frac{3}{4}\times \frac{1}{4}+\frac{1}{2}\times \frac{1}{4}\times \frac{3}{4}}{\frac{1}{4}}=\frac{7}{8}$ $\therefore$ (b) is true. (c) $P\left(X/X_2\right)=\frac{P\left(X\cap X_2\right)}{P\left(X_2\right)}$ $=\frac{P\left(X_1\cap X_2\cap X_3\right)+P\left(X_1^C\cap X_2\cap X_3\right)+P\left(X_1\cap X_2\cap X_3^C\right)}{P\left(X_2\right)}$ $=\frac{\frac{1}{2}\times \frac{1}{4}\times \frac{1}{4}+\frac{1}{2}\times \frac{1}{4}\times \frac{1}{4}+\frac{1}{2}\times \frac{1}{4}\times \frac{3}{4}}{\frac{1}{4}}=\frac{5}{8}$ $\therefore$ (c) is not true. (d) $P\left(X/X_1\right)=\frac{P\left(X\cap X_1\right)}{P\left(X_1\right)}$ $=\frac{P\left(X_1\cap X_2\cap X_3\right)+P\left(X_1\cap X_2^C\cap X_3\right)+P\left(X_1\cap X_2\cap X_3^C\right)}{P\left(X_1\right)}$ $=\frac{\frac{1}{2}\times \frac{1}{4}\times \frac{1}{4}+\frac{1}{2}\times \frac{3}{4}\times \frac{1}{4}+\frac{1}{2}\times \frac{1}{4}\times \frac{3}{4}}{\frac{1}{2}}=\frac{7}{16}$ $\therefore$ (d) is true.