Question

Shubham draws a card from a pack of cards, replaces it, shuffles the pack and then draws another card. What is the probability that the cards are both aces?

• A. $\frac{1}{169}$
• B. $\frac{1}{196}$
• C. $\frac{1}{144}$
• D. $\frac{1}{13}$

Answer 1

Answer: (A)

$\frac{1}{169}$

$P\left(ace\right)=\frac{4}{52}=\frac{1}{13}$ ($\because$ there are 4 pack of 52 cards)
$\therefore P\left(bothaces\right)=P\left(firstaceandsecondace\right)=\frac{1}{13}\times \frac{1}{13}=\frac{1}{169}$