# Deck of Cards Probability question?

Three cards are drawn from a deck of 52 playing cards and not replaced. Find the probability of the following in fraction form.

a) Getting three jacks

b) Getting an ace, a king, and a queen in order.

c) Getting a club, a spade, and a heart in order

d) Getting three clubs)

### 4 Answers

- David TLv 41 decade agoFavorite Answer
(Probability of first draw)*(prob of 2nd draw)*(prob of 3rd draw)

a) (4/52)*(3/51)*(2/50)

b) (4/52)*(4/51)*(4/50)

c) (13/52)*(13/51)*(13/50)

d) (13/52)*(12/51)*(11/50)

The denominator in each case is the number of cards remaining in the deck, and so reduces by one each time you draw a card (some of the fractions can be simplified, but I left it unsimplified so you can see the relation to the number of cards). The numerator is the number of cards remaining in the deck for that draw that match the criteria.

- 1 decade ago
4 jacks: 4/52 * 3/51 * 2/50 = 1/5525

4 aces, kings, queens: 4/52 * 4/51 * 4/50 = 8/16575

13 clubs, 13 spades, 13 hearts: 13/52 * 13/51 * 13/50 = 169/10200

13 clubs: 13/52 * 12/51 * 11/50 = 11/850

Hope this helps! :)

- 1 decade ago
Okay, 3 out of 52 could be selected in 52 choose 3 ways=2210

1)3 out of 4 Jacks could be selected in 4 choose 3 ways.

Probability of event (a) is 4/2210

2)An ace, a king and queen in this order could be selected in 4x4x4=64 ways.

Probability of event (b) is 64/2210

3)A club, a spade and a heart(including face cards in each) in this order in 13x13x13 ways=2197 ways.

Probabilty of event (c)=2197/2210

4)3 clubs(including face cards) could occur in 13choose 3 ways=286 ways.

Probability of event (d) is 286/2210

- ag_iitkgpLv 71 decade ago
a) 4C3 / 52C3 = 1/11050

b) 4/52 x 4/51 x 4/50 = 8/16575

c) 13/52 x 13/51 x 13/50 = 169/10200

d) 13C3 / 52C3 = (13x12x11)/(50x51x52) = 11/850