Question

Two cards are drawn from a well shuffled pack of 52 playing cards one by one. If
$A$: the event that the second card drawn is an ace and
$B$: the event that the first card drawn is an ace card.
then which of the following is true?

• A. $P\left(A\right)=\frac{4}{17};P\left(B\right)=\frac{1}{13}$
• B. $P\left(A\right)=\frac{1}{13};P\left(B\right)=\frac{1}{13}$
• C. $P\left(A\right)=\frac{1}{13};P\left(B\right)=\frac{1}{17}$
• D. $P\left(A\right)=\frac{16}{221};P\left(B\right)=\frac{4}{51}$

Answer 1

The correct option is B $P\left(A\right)=\frac{1}{13};P\left(B\right)=\frac{1}{13}$

Given, among 52 pack of cards 2 are drawn. We know that in a pack of cards,there will be 4 aces.
1)Probability of second drawn card is an ace$=$Probability of $first$ drawn card is $ace$ and $second$ drawn is an $ace$$+$probability of $first$ drawn is $non-ace$ and $second$ drawn is as $ace$
$=\frac{4}{52}\ast \frac{3}{51}+\frac{48}{52}\ast \frac{4}{51}=(\frac{1}{13}\ast \frac{1}{17})+\left(\frac{16}{13\ast 17}\right)=\frac{1}{13}$

2)probaility of first drawn is ace$=$$\frac{4}{52}\ast \frac{51}{51}=\frac{1}{13}$ (since we are not worrying about second card,it can be anything among $51$cards)