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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
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Verified Answer
The correct answer is:
5 or 10
5 or 10
$P(A)=\frac{2}{5}$
For independent events,
$$
\begin{aligned}
& P(A \cap B)=P(A) P(B) \Rightarrow P(A \cap B) \leq \frac{2}{5} \\
& \Rightarrow P(A \cap B)=\frac{1}{10}, \frac{2}{10}, \frac{3}{10}, \frac{4}{10} \\
& \text { (i) } P(A \cap B)=\frac{1}{10} \\
& \Rightarrow P(A) \cdot P(B)=\frac{1}{10} \\
& \Rightarrow \quad P(B)=\frac{1}{10} \times \frac{5}{2}=\frac{1}{4}, \text { not possible. } \\
& \text { (ii) } P(A \cap B)=\frac{2}{10} \Rightarrow \frac{2}{5} \times P(B)=\frac{2}{10} \\
& \Rightarrow \quad P(B)=\frac{5}{10}, \text { outcomes of } B=5
\end{aligned}
$$
$$
\text { (iii) } \begin{aligned}
& P(A \cap B)=\frac{3}{10} \Rightarrow P(A) \cdot P(B)=\frac{3}{10} \\
& \Rightarrow \quad \frac{2}{5} \times P(B)=\frac{3}{10} \\
& P(B)=\frac{3}{4}, \text { not possible }
\end{aligned}
$$
$$
\text { (iv) } \begin{aligned}
& P(A \cap B)=\frac{4}{10} \\
\Rightarrow & P(A) \cdot P(B)=\frac{4}{10} \\
\Rightarrow & P(B)=1, \text { outcomes of } B=10 .
\end{aligned}
$$
For independent events,
$$
\begin{aligned}
& P(A \cap B)=P(A) P(B) \Rightarrow P(A \cap B) \leq \frac{2}{5} \\
& \Rightarrow P(A \cap B)=\frac{1}{10}, \frac{2}{10}, \frac{3}{10}, \frac{4}{10} \\
& \text { (i) } P(A \cap B)=\frac{1}{10} \\
& \Rightarrow P(A) \cdot P(B)=\frac{1}{10} \\
& \Rightarrow \quad P(B)=\frac{1}{10} \times \frac{5}{2}=\frac{1}{4}, \text { not possible. } \\
& \text { (ii) } P(A \cap B)=\frac{2}{10} \Rightarrow \frac{2}{5} \times P(B)=\frac{2}{10} \\
& \Rightarrow \quad P(B)=\frac{5}{10}, \text { outcomes of } B=5
\end{aligned}
$$
$$
\text { (iii) } \begin{aligned}
& P(A \cap B)=\frac{3}{10} \Rightarrow P(A) \cdot P(B)=\frac{3}{10} \\
& \Rightarrow \quad \frac{2}{5} \times P(B)=\frac{3}{10} \\
& P(B)=\frac{3}{4}, \text { not possible }
\end{aligned}
$$
$$
\text { (iv) } \begin{aligned}
& P(A \cap B)=\frac{4}{10} \\
\Rightarrow & P(A) \cdot P(B)=\frac{4}{10} \\
\Rightarrow & P(B)=1, \text { outcomes of } B=10 .
\end{aligned}
$$
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