Probability Ques 93
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
Answer:
Correct Answer: 93.(d)
Solution:
Formula:
- Since, $P(A)=\frac{2}{5}$
For independent events,
$P(A \cap B)=P(A) P(B) $
$\Rightarrow \quad P(A \cap B) \leq \frac{2}{5} $
$\Rightarrow \quad P(A \cap B)=\frac{1}{10}, \frac{2}{10}, \frac{3}{10}, \frac{4}{10}$
[maximum $4 $ outcomes may be in $A \cap B$ ]
(i) Now, $\quad P(A \cap B)=\frac{1}{10}$
$\Rightarrow P(A) \cdot P(B)=\frac{1}{10} $
$\Rightarrow P(B)=\frac{1}{10} \times \frac{5}{2}=\frac{1}{4}, \text { not possible }$
(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, $\quad P(A \cap B)=\frac{3}{10}$
$\Rightarrow \quad P(A) P(B)=\frac{3}{10} \Rightarrow \frac{2}{5} \times P(B)=\frac{3}{10} $
$P(B)=\frac{3}{4}, \text { not possible }$
(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 {. } $
BETA
