Explanation

(d)
$P(A) = \frac{2}{5}$
For independent events,
$P(A \cap B) = P(A)P(B$)
$\Rightarrow P(A \cap B) \leq \frac{2}{5}$
$\Rightarrow P(A \cap B) = \frac{1}{10},\frac{2}{10},\frac{3}{10},\frac{4}{10}$
[Maximum 4 outcomes may be in $P(A \cap B)\rbrack$
1. When$P(A \cap B) = \frac{1}{10}$
$\Rightarrow P(A).P(B) = \frac{1}{10}$
$\Rightarrow P(B) = \frac{1}{10} \times \frac{5}{2} = \frac{1}{4},$not possible
2. When $P(A \cap B) = \frac{2}{10} \Rightarrow \frac{2}{5} \times P(B) = \frac{2}{10}$
$\Rightarrow P(B) = \frac{5}{10},$outcomes of $B = 5$
3. When $P(A \cap B) = \frac{3}{10}$
$\Rightarrow P(A)P(B) = \frac{3}{10}$
$\Rightarrow \frac{2}{5} \times P(B) = \frac{3}{10}$
$P(B) = \frac{3}{4},$not possible
4. When $P(A \cap B) = \frac{4}{10}$
$\Rightarrow P(A).P(A) = \frac{4}{10}$
$\Rightarrow P(B) = 1,\$outcomes of $B = 10$