Question

Next year there will be three candidates Mr.X, Mr.Y and Mr.Z for the position of a principal of a degree college exclusively meant for boys. Their respective chances for the selection are in proportion $4:2:3$. The probabilities that these persons, if selected will introduce co-education in the college, are respectively $0.3,0.5$ and $0.8$. The probability, that there will be co-education in the college, next year is $\frac{23}{9K}$, then $K$ is

Answer 1

$\frac{P\left(X\right)}{4}=\frac{P\left(Y\right)}{2}=\frac{P\left(Z\right)}{3}=\lambda$(suppose)
Then,$P\left(X\right)+P\left(Y\right)+P\left(Z\right)=1\Rightarrow 4\lambda +2\lambda +3\lambda =1\therefore \lambda =\frac{1}{9}$
$P\left(X\right)=\frac{4}{9},P\left(Y\right)=\frac{2}{9}$and$P\left(Z\right)=\frac{3}{9}$
Let '$E$' be the event of introducing coeducation. Using total probability theorem, we have:
$P\left(E\right)=\sum P\left(X\right).P\left(E/X\right)=[\frac{4}{9}\times \frac{3}{10}+\frac{2}{9}\times \frac{5}{10}+\frac{3}{9}\times \frac{8}{10}]=\frac{46}{90}=\frac{23}{45}$
Then $9K=45\Rightarrow K=5$