Question

Mark the correct alternative in the following question:

Two cards are drawn from a well shuffled deck of 52 playing cards with replacement. The probability that both cards are queen is

$\left(\mathrm{a}\right)\frac{1}{13}\times \frac{1}{13}\left(\mathrm{b}\right)\frac{1}{13}+\frac{1}{13}\left(\mathrm{c}\right)\frac{1}{13}\times \frac{1}{17}\left(\mathrm{d}\right)\frac{1}{13}\times \frac{4}{5}$

Answer 1

$$\begin{gathered} \mathrm{Let}: \\ A\mathrm{be}\mathrm{the}\mathrm{event}\mathrm{that}\mathrm{a}\mathrm{queen}\mathrm{is}\mathrm{drawn}\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{draw}\mathrm{and} \\ B\mathrm{be}\mathrm{the}\mathrm{event}\mathrm{that}\mathrm{a}\mathrm{queen}\mathrm{is}\mathrm{drawn}\mathrm{in}\mathrm{the}\mathrm{second}\mathrm{draw}\mathrm{as}\mathrm{well} \\ \mathrm{Now}, \\ \mathrm{P}\left(\mathrm{Both}\mathrm{the}\mathrm{two}\mathrm{cards}\mathrm{drawn}\mathrm{are}\mathrm{queen}\right)=\mathrm{P}\left(A\right)\times \mathrm{P}\left(B|A\right) \\ =\frac{4}{52}\times \frac{4}{52} \\ =\frac{1}{13}\times \frac{1}{13} \end{gathered}$$

Hence, the correct alternative is option (a).