Two cards are drawn at random from a pack of 52 cards without replacem

Question

Two cards are drawn at random from a pack of 52 cards without replacement. What is the probability that one is a club and other is an ace?

A. 1/52
B. 1/26
C. 1/13
D. 4/13
E. 7/13

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GMATGuruNY · Accepted Answer

Alternate approach:

AND means MULTIPLY.
OR means ADD.

Case 1: P(ace of spades, diamonds or hearts AND any type of clubs) 
P(first card is an ace of spades, diamonds or hearts) = 3/52 (Of the 52 cards, 3 are the ace of spades, diamonds or hearts)
P(second card is clubs) = 13/51 (Of the 51 remaining cards, 13 are clubs)
Since Case 1 requires the first card AND the second card, we multiply:
3/52 * 13/51 = 1/4 * 1/17
Since the two cards can also be picked in reverse order, we multiply by 2:
1/4 * 1/17 * 2 =  1/2 * 1/17

Case 2: P(ace of clubs AND any other type of clubs) 
P(first card is ace of clubs) = 1/52 (Of the 52 cards, 1 is the ace of clubs)
P(second card is clubs) = 12/51 (Of the 51 remaining cards, 12 are clubs)
Since Case 2 requires the first card AND the second card, we multiply:
1/52 * 12/51 = 1/13 * 1/17
Since the two cards can also be picked in reverse order, we multiply by 2:
1/13 * 1/17 * 2 =  2/13 * 1/17

Since a favorable outcome will be yielded by Case 1 OR Case 2, we add the results in blue:
(1/2 * 1/17) + (2/13 +1/17) = (1/17)(1/2 + 2/13) = (1/17)(17/26) = 1/26

B

chetan2u · Answer

We have 13 clubs cards, out of which one is an Ace.
Total ways to select 2 cards out of 52 =  52C2=52*\frac{51}{2}=26*51

Ways to select the given scenario 
1) Any of 3 aces and one of the 13 clubs(including the club ace) =  3C1*13C1=3*13=39 
2) Ace of club and one out of the remaining 12 club cards =  1C1*12C1=1*12=12 
Total = 39+12=51

P= \frac{51}{26*51}=\frac{1}{26}

B

MBAApplicant2026 · Answer

Why do we not consider the case where we can select the ace of clubs and also another ace? Wouldn’t that satisfy the condition of choosing an ace and a club?

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chetan2u · Answer

The above is already included in 
Any of the remaining three aces and any of club(incl ACE club)
3C1*13C1 = 3*13 = 39

Regor60 · Answer

So it's clear that the tricky part is the Ace of clubs.

The thing to realize is that the Ace of clubs can fill only one role at a time, as an ace or as a club, but not both at the same time since it's clear from the wording of the question that they need to be represented on different cards.

Two cards are drawn at random from a pack of 52 cards without replacem

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