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In a box, there are $20$ cards, out of which $10$ are lebelled as $\mathrm{A}$ and the 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
$\frac{11}{16}$
$\frac{13}{16}$
$\frac{9}{16}$
$\frac{15}{16}$
Solution
$A:$ Event when card $A$ is drawn
$\mathrm{B}:$ Event when card $\mathrm{B}$ is drawn.
$\mathrm{P}(\mathrm{A})=\mathrm{P}(\mathrm{B})=\frac{1}{2}$
Required probability $=\mathrm{P}(\mathrm{AA} \text { or }(\mathrm{AB}) \mathrm{A}$
or $(\mathrm{BA}) \mathrm{A} \text { or }(\mathrm{ABB}) \mathrm{A} \text { or }(\mathrm{BAB}) \mathrm{A} \text { or }(\mathrm{BBA}) \mathrm{A})$
$=\frac{1}{2} \times \frac{1}{2}+\left(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\right) \times 2+\left(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\right) \times 3$
$=\frac{1}{4}+\frac{1}{4}+\frac{3}{16}=\frac{11}{16}$
