Question

In a box, there are $20$ cards out of which $10$ are labelled as $A$ and remaining $10$ are labelled as $B$. Cards are drawn at random, one after the other and with replacement, till a second $A$-card is obtained. The probability that the second $A$ -card appears before the third $B$-card is :

• A. $\frac{15}{16}$
• B. $\frac{9}{16}$
• C. $\frac{13}{16}$
• D. $\frac{11}{16}$

Answer 1

The correct option is D $\frac{11}{16}$

Here, $P\left(A\right)=P\left(B\right)=\frac{1}{2}$
Then, these following cases are possible $\to$ $AA,BAA,ABA,ABBA,BBAA,BABA$

So, the required probability is $=\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}$
$=\frac{11}{16}$