Probability 3 Question 9
9. 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, then the number of outcomes that $B$ must have, so that $A$ and $B$ are independent, is
(a) 2,4 or 8
(b) 3,6 or 9
(c) 4 or 8
(d) 5 or 10
$(2008,3$ M)
Show Answer
Solution:
- Since, $P(A)=\frac{2}{5}$
For independent events,
$$ \begin{aligned} & P(A \cap B)=P(A) P(B) \\ \Rightarrow & P(A \cap B) \leq \frac{2}{5} \\ \Rightarrow \quad & P(A \cap B)=\frac{1}{10}, \frac{2}{10}, \frac{3}{10}, \frac{4}{10} \end{aligned} $$
[maximum 4 outcomes may be in $A \cap B$ ]
(i) Now, $\quad P(A \cap B)=\frac{1}{10}$
$$ \begin{array}{ll} \Rightarrow & P(A) \cdot P(B)=\frac{1}{10} \\ \Rightarrow & P(B)=\frac{1}{10} \times \frac{5}{2}=\frac{1}{4}, \text { not possible } \end{array} $$
(ii) Now, $\quad P(A \cap B)=\frac{2}{10} \Rightarrow \frac{2}{5} \times P(B)=\frac{2}{10}$
$\Rightarrow \quad P(B)=\frac{5}{10}$, outcomes of $B=5$
(iii) Now, $\quad P(A \cap B)=\frac{3}{10}$
$$ \begin{gathered} \Rightarrow \quad P(A) P(B)=\frac{3}{10} \Rightarrow \frac{2}{5} \times P(B)=\frac{3}{10} \\ P(B)=\frac{3}{4}, \text { not possible } \end{gathered} $$
(iv) Now,
$$ P(A \cap B)=\frac{4}{10} \Rightarrow P(A) \cdot p(B)=\frac{4}{10} $$
$$ \Rightarrow \quad P(B)=1 \text {, outcomes of } B=10 \text {. } $$
