An experiment has 10 equally likely outcomes. Let  $A$ and  $B$ be two non-empty events of the experiment. If  $A$ consists of 4 outcomes, the number of outcomes that  $B$ must have so that $A$  and  $B$ are independent, is

$P\left(A\right)=\frac{2}{5}$

For independent events

$$\begin{gathered} P(A\cap B)=P\left(A\right)P\left(B\right) \\ P(A\cap B)\le \frac{2}{5} \\ \Rightarrow P(A\cap B)=\frac{1}{10},\frac{2}{10},\frac{3}{10},\frac{4}{10}(\text{Maximum\hspace{0.17em}4\hspace{0.17em}outcomes\hspace{0.17em}may\hspace{0.17em}be\hspace{0.17em}in\hspace{0.17em}}A\cap B) \\ i) P(A\cap B)=\frac{1}{10} \\ \Rightarrow P\left(A\right).P\left(B\right)=\frac{1}{10} \\ P\left(B\right)=\frac{1}{10}\times \frac{5}{2}=\frac{1}{4},\text{\hspace{0.17em}not possible} \end{gathered}$$

$$\begin{gathered} ii) P(A\cap B)=\frac{2}{10} \\ \Rightarrow \frac{2}{5}\times P\left(B\right)=\frac{2}{10} \\ \Rightarrow P\left(B\right)=\frac{5}{10},\text{\hspace{0.17em}outcomes\hspace{0.17em}of\hspace{0.17em}}B=5 \\ iii) P(A\cap B)=\frac{3}{10} \\ P\left(A\right)P\left(B\right)=\frac{3}{10} \\ \frac{2}{5}\times P\left(B\right)=\frac{3}{10} \\ P\left(B\right)=\frac{3}{4},\text{\hspace{0.17em}not\hspace{0.17em}possible} \end{gathered}$$

$$\begin{gathered} iv) P(A\cap B)=\frac{4}{10} \\ \Rightarrow P\left(A\right).P\left(B\right)=\frac{4}{10} \\ \Rightarrow P\left(B\right)=1,\text{\hspace{0.17em}outcomes\hspace{0.17em}of\hspace{0.17em}}B=10 \end{gathered}$$