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# Two cards are drawn from a regular deck of 52 cards, without replacement.?

Two cards are drawn from a regular deck of 52 cards, without replacement. What is the probability that the first card is an ace of clubs and the second is black?

Help please!!! This is Honors Algebra 3-4

### 6 Answers

- lhvinnyLv 71 decade agoFavorite Answer
There is only one ace of clubs in the deck, so the probability for the first event is 1/52.

After removal of the ace of clubs, there are 25 black cards in the deck (13 spades + all the clubs except for the ace). That means a 25/51 probability.

In a combined probability, the event probabilities are multiplied:

1/52 * 25/51 = 25 / 2652

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- i + iLv 71 decade ago
First draw would be simply 1 in 52 chance of it being the ace of clubs. One assumes that the first event actually occured (an ace of clubs WAS drawn) in computing the probability for the second card -- so the probabilities for the second draw would therefore be 25 in 51 chances of it being black. You multiply these together to get the overall probability (ace of clubs followed by a black card): 1/52*25/51 = 25/2652

[edited to fix my horrible English]

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- 4 years ago
Probability that the first card is an ace of clubs: 1/52 (only one ace of clubs) Probability that the second card is black: 25/51 (there are only 51 cards left and now only 25 black cards) So the probability of both of those is: 1/52 * 25/51

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- 1 decade ago
The chance of the first being an A♣ is 1/52.

The chance of the second being black is 25/51.

Multiply to get 25/2652, or about 0.9427%.

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- Anonymous1 decade ago
1/52 x 25/51

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- 1 decade ago
1/52*25/51

that's not very hard. I'm doing algebra 2 and I knew it...

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