Probability: For these problems, consider a standard American deck of playing cards-52 cards, 4 suits in the deck, 13 cards in each suit. Imagine also that the deck has been well shuffled.

1. Draw 1 card. What is the probability of drawing a diamond?

2. Draw 2 cards without replacement. What is the probability that both are diamonds?

3. Draw 2 cards without replacement. What is the probability that the first is a diamond but the second is a club?

4. Draw 2 cards without replacement. What is the probability that the first is a club but the second is a diamond?

5. Draw 2 cards without replacement. What is the probability that one of them is a diamond and the other is a club?

For all probability exercises, the probability is always

(the number of possible 'successful' results)

divided by

(the total possible number of results).

1). Number of diamonds in the deck = 13

Total cards in the deck = 52.

Probability = (13/52) = 1/4 = 25%

2). Probability of the first one being a diamond = 1/4.

Now there are 51 cards left, and 12 of them are diamonds.

Probability of the second one being a diamond = 12/51

Probability of both events =

(1/4) x (12/51) = 3/51 = about 5.88% (rounded)

3). First one is a diamond: 13/52

Now there are 51 cards left, and 13 of them are clubs.

Second one is a club: 13/51.

Probability of both events = (13/52) x (13/51)

= 169/2652 = about 6.37% (rounded)

4). Same logic, same solution, as #3.

5). 6.37% chance that it happens with the diamond coming up first,

6.37% chance that it happens with the club coming up first.

Probability of either one happening = 12.74% (rounded).