Question

Mark the correct alternative in the following question:

$$\begin{gathered} \mathrm{If}A\mathrm{and}B\mathrm{are}\mathrm{two}\mathrm{independent}\mathrm{events}\mathrm{with}\mathrm{P}\left(A\right)=\frac{3}{5}\mathrm{and}\mathrm{P}\left(B\right)=\frac{4}{9},\mathrm{then}\mathrm{P}\left(\overline{A}\cap \overline{B}\right)\mathrm{equals} \\ \left(\mathrm{a}\right)\frac{4}{15}\left(\mathrm{b}\right)\frac{8}{45}\left(\mathrm{c}\right)\frac{1}{3}\left(\mathrm{d}\right)\frac{2}{9} \end{gathered}$$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{P}\left(A\right)=\frac{3}{5}\mathrm{and}\mathrm{P}\left(B\right)=\frac{4}{9} \\ \mathrm{As},A\mathrm{and}B\mathrm{are}\mathrm{independent}\mathrm{events} \\ \mathrm{So},\mathrm{P}\left(A\cap B\right)=\mathrm{P}\left(A\right)\times \mathrm{P}\left(B\right) \\ =\frac{3}{5}\times \frac{4}{9} \\ =\frac{12}{45} \\ =\frac{4}{15} \\ \mathrm{Also},\mathrm{P}\left(A\cup B\right)=\mathrm{P}\left(A\right)+\mathrm{P}\left(B\right)-\mathrm{P}\left(A\cap B\right) \\ =\frac{3}{5}+\frac{4}{9}-\frac{4}{15} \\ =\frac{27+20-12}{45} \\ =\frac{35}{45} \\ =\frac{7}{9} \\ \mathrm{Now}, \\ \mathrm{P}\left(\overline{A}\cap \overline{B}\right)=\mathrm{P}\left(\overline{A\cup B}\right) \\ =1-\mathrm{P}\left(A\cup B\right) \\ =1-\frac{7}{9} \\ =\frac{2}{9} \end{gathered}$$

Hence, the correct alternative is option (d).