Question

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. $5 \,or\, 10$

Answer 1

Answer: (D)

$5 \,or\, 10$

Answer 2

Since, P(A) =$\frac{2}{5}$
For independent events,
P(A $\cap$B) = P(A)P(B) $\Rightarrow \, P(A\cap B) \le \frac{2}{5}$
$\Rightarrow \, \, \, P(A \cap B)=\frac{1}{10},\frac{2}{10},\frac{3}{10},\frac{4}{10}$
[maximum 4 outcomes may be in $A \cap B$
(i) Now, $\, \, \, \, \, \, P(A \cap B)=\frac{1}{10}$
$\Rightarrow \, \, \, \, \, \, \, \, P(A).P(B) = \frac{1}{10}$
$\Rightarrow \, \, \, \, \, \, \, \, P(B)=\frac{1}{10} \times \frac{5}{2}=\frac{1}{4},$ not possible
(ii) Now,$\, \, \, \, \, \, \, \, \, \, \, P(A \cap B)=\frac{2}{10} \, \, \Rightarrow \, \, \frac{2}{5} \times P(B)=\frac{2}{10}$
$\Rightarrow \, \, \, \, \, \, \, \, \, P(B)=\frac{5}{10},$outcomes of B = 5
(iii) Now, $P(A \cap B)=\frac{3}{10}$
$\Rightarrow \, \, \, \, \, \, \, \, \, P(A)P(B)=\frac{3}{10} \, \, \Rightarrow \, \, \, \frac{2}{5}\times P(B) =\frac{3}{10}$
P(B) =$\frac{3}{4},$ not possible
(iv) Now $P(A \cap B)=\frac{4}{10} \Rightarrow \, \, P(A)>p(B)=\frac{4}{10}$
$\Rightarrow \, \, \, \, \, \, \, \, \, \, \, P(B)=1,$ outcomes of B = 10.