Question

Sara selects two cards at random from a standard deck of fifty-two cards. Which of the following could be used to calculate the probability that she will select two numbered cards if she does not replace the first card before selecting the second? Note: For this problem, face cards and aces are not numbered cards.

thirty-six divided by fifty-two times thirty-five divided by fifty-one

thirty-six divided by fifty-two times thirty-five divided by fifty-two

thirty-six divided by fifty-two times thirty-six divided by fifty-two

thirty-six divided by fifty-two times thirty-six divided by

Answer 1

1. None of these works for the problem as you have described it. If you had said "odd" instread
of "even", then B would be your answer. For picking odd numbers, you'd have the
3, 5, 7, and 9 each of which occurs 4 times for a total of 16 good picks out of 52 cards.
This gets you 16/52 for the first fraction. After picking 1 odd card without replacement you would
then have 15 odd cards left in a deck of 51 cards, which gets you the 15/51 for the 2nd fraction.

The reason that this doesn't work for "even" is that you have 2, 4, 6, 8, 10. So the numbers
on the top of the fractions would have to be 20 and 19.

Answer 2

There are 36 numbered cards. On the first selection the probability of selecting a numbered card is 36/52. Given that a numbered card has been selected first , the probability of getting another numbered card on the second selection is 35/51.
therefore the required probability is given by:
\[P(2\ numbered\ cards)=\frac{36}{52}\times\frac{35}{51}\]