To solve this question, since Peterson is placing the card back in the deck, we have to compute the probability of getting a Jack from a deck of cards, then the probability of getting a 10 numbered card from a deck of cards. The answer is the product of the probabilities.
The probability of getting a Jack, since there are 4 Jacks, is:
[tex]\frac{4}{52}\text{.}[/tex]The probability of getting a 10 numbered card, since there are 4 cards of this type, is:
[tex]\frac{4}{52}\text{.}[/tex]Therefore, the probability of choosing a Jack and then a 10 numbered card is:
[tex]\frac{4}{52}\times\frac{4}{52}=\frac{1}{169}\text{.}[/tex]Answer: Second option.
