what is the probability of selecting two clubs when two cards are drawn(without replacment) from a standard deck of cards?

Question

Not sure, i've been having trouble

Answer 1

Let A be the event "the first card drawn is club" and B be the event "the second card drawn is club". We want to know the probability that both A and B occur. Since the drawing is without replacement, A and B are dependent events, so use the multiplication rule for dependent events:

P(A and B) = P(A)*P(B|A) = (13/52)*(12/51) = 1/17.

Alternatively, let the random variable X count the number of clubs in two draws. Then X is hypergeometric with population size 52, sample size 2, and 13 successes in the population. The desired probability is given by the hypergeometric pmf:

P(X=2) = (13C2)(39C0)/(52C2) = 1/17.

Answer 2

Hi Aaron

The probability of drawing the first club is 13/52. Without replacement, there are 12 clubs and 51 total cards, so the probability of drawing the second club is 12/51.

Since these are independent events, the probability of two clubs is the product of these two probabilities.

13/52 * 12/51 = 3/51 = .059