Question

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

• A. $\frac{15}{6}$
• B. $\frac{9}{16}$
• C. $\frac{13}{16}$
• D. $\frac{11}{16}$

Answer 1

The correct option is D

$\frac{11}{16}$

Explanation for the correct option:

Finding the probability:

Given that

Total no. of cards $=20$

No. of cards labeled as $\mathrm{A}$ $=10$

No of cards labeled as $\mathrm{B}$$=10$

$\mathrm{P}\left(\mathrm{A}\right)=\mathrm{P}\left(\mathrm{B}\right)=\frac{1}{2}$

Now, according to question, these are the following arrangements or cases of cards that can appear from the bundle,

$=\mathrm{AA},\mathrm{BAA},\mathrm{ABA},\mathrm{ABBA},\mathrm{BBAA},\mathrm{BABA}$

Hence, the probability of getting second $\mathrm{A}$ before getting a third $\mathrm{B}$$=\mathrm{P}\left(\mathrm{AA}\right)+\mathrm{P}\left(\mathrm{BAA}\right)+\mathrm{P}\left(\mathrm{ABA}\right)+\mathrm{P}\left(\mathrm{ABBA}\right)+\mathrm{P}\left(\mathrm{BBAA}\right)+\mathrm{P}\left(\mathrm{BABA}\right)$

$$\begin{gathered} =\left(\mathrm{P}\right(\mathrm{A})\times \mathrm{P}(\mathrm{A}\left)\right)+\left(\mathrm{P}\right(\mathrm{B})\times \mathrm{P}(\mathrm{A}\left)\mathrm{P}\right(\mathrm{A}\left)\right)+\left(\mathrm{P}\right(\mathrm{A})\times \mathrm{P}(\mathrm{B})\times \mathrm{P}(\mathrm{A}\left)\right)+\left(\mathrm{P}\right(\mathrm{A})\times \mathrm{P}(\mathrm{B}) \\ \times \mathrm{P}(\mathrm{B})\times \mathrm{P}(\mathrm{A}\left)\right)+\left(\mathrm{P}\right(\mathrm{B})\times \mathrm{P}(\mathrm{A})\times \mathrm{P}(\mathrm{B})\times \mathrm{P}(\mathrm{A}\left)\right)+\left(\mathrm{P}\right(\mathrm{B})\times \mathrm{P}(\mathrm{B})\times \mathrm{P}(\mathrm{A})\times \mathrm{P}(\mathrm{A}\left)\right) \end{gathered}$$

$=\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}$

$=\frac{11}{16}.$

Hence the correct option is D.