What is the probability of getting an ace on the first card and a spade on the second card (no replacement)?

Question

Drawing an ace would be 4/52. Then drawing a spade would be 13/51? but then i considered the possibility of getting an ace of spade but i dont know how to take that into account... I'm stuck @_@

The answer is .0192. Please explain how to do it.

Thanks so much in advance!

Answer 1

You're right its 2 seperate cases:
First no ace spade on first card
3/52 * 13/51

Then drawing ace spade on first card
1/52 * 12/51.

Just add the two probabilities together to get the answer you posted, which conveniently is 1/52.

Answer 2

There are 52*51 =2652 ways of dealing 2 cards.

There are 13+13+13+12 =51 ways of getting an Ace and then a spade. The first three 13's is if the Ace was not a spade, and the 12 is when the Ace was a spade.

So the odds are 51/2652 =1/52 = approx 0.0192

Answer 3

The probability of drawing an ace and drawing a spade are independent so you can just multiply the probabilties together.

P(A then Spade) = (1/13)(1/4) = 1/52 ≈ 0.0192

Answer 4

probability of drawing ace of spades on first card = 1/52
since you mentioned no replacement..
probability to get a spade on second card = 12/51

Answer 5

(1/4)(1/13)(12/51)
+(3/4)(1/13)(13/51) = 0.01923
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Ideas: Partition the case into two.
Case I: The first card is a spade.(p = 1/4)
Case II: The first card is not a spade.(p = 3/4)

Answer 6

P(Ace & spade) = (3/52)(13/51) + (1/52)(12/51) =
(39 + 12) / (51)(52) = 51 / (51)(52) ≈ 0.01923

Answer 7

when you somethings with gether you can use * in calculate
and use + , when you want one of them

4/52 * 13/51