Mathematics - Probability Question with Solution | TestHub

Let $E=$ event when each American man is seated adjacent to his wife and $A=$ event when Indian man is seated adjacent to his wife. Now, $n(A\cap E)=(4!)\times (2!{)}^5$ Even when each American man is seated adjacent to his wife. Again $$\begin{array}{rl}n(E)& =(5!)\times (2!{)}^4\\ P\left(\frac{A}{E}\right)& =\frac{n(A\cap E)}{n(E)}\\ & =\frac{(4!)\times (2!{)}^5}{(5!)\times (2!{)}^4}=\frac{2}{5}\end{array}$$ $$\Rightarrow P\left(\frac{A}{E}\right)=\frac{n(A\cap E)}{n(E)}$$ ALITER Fixing four American couples and one Indian man in between any two couples, we have 5 different ways in which his wife can be seated, of which 2 cases are favourable. $\therefore$ Required probability $=\frac{2}{5}$.