Your solution is correct. However, it could be reduced to two cases.
- The first card is a club that is not the ace of clubs and the second card is an ace.
- The first card is the ace of clubs and the second card is an ace that is not the ace of clubs.
Case 1: There are twelve ways of selecting a club that is not the ace of clubs in the first draw and four ways to select one of the four available aces in the second draw. Hence, there are $12 \cdot 4 = 48$ possible selections in the this case.
Case 2: There is one way to select the ace of clubs in the first draw and three ways to select one of the three remaining aces in the second draw. Hence, there are $1 \cdot 3 = 3$ possible selections in the second case.
Total: $12 \cdot 4 + 1 \cdot 3 = 48 + 3 = 51$
The authors of your text seem to be answering a different question, namely, in how many ways can exactly one ace and exactly one club be drawn if two cards are selected from a deck?
They do not take into account the fact that the first card is a club and the second card is an ace. They are allowing the cards to be drawn in either order.
Ace of clubs is one of those cards:
Select the ace of clubs, select one of the $36$ cards that is neither an ace nor a club, then multiply by $2$ to account for the two possible orders in which the cards could be selected.
Ace of clubs is not one of those cards:
Select one of the other three aces, select one of the twelve clubs that is not the ace of clubs, then multiply by $2$ to account for the two possible orders in which the cards could be selected.
2 Answers By Expert Tutors
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.
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
Two cards are drawn together from a pack of 52 cards. The probability that one is a club and one is a diamond ?
n(S) = Total number of ways of drawing 2 cards from 52 cards = 52C2. Let E = event of getting 1 club and 1 diamond. We know that there are 13 clubs and 13 diamonds in the total 52 cards.
Hence, n(E) = Number of ways of drawing one club from 13 and one diamond from 13 = 13C1 × 13C1 = 13/102
P(E) = n(E)/n(S)= (13C1 × 13C1) / 52C2
= 13/102.
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