1. ## Conditional Probability

I have some ambiguity in the following problem
Three cards are drawn in succession from a deck without replacement. find the probability distribution for the number of spades.
I have come with this solution.
Let S1: appearance of spade on first draw S2: appearance of spade on 2nd draw S3: appearance of spade on 3rd draw
S01: No spade on 1st draw S02: No spade on 2nd draw
P(S1)=13/52
P(S2)=P(S2/S1)+P(S2/S0)=12/51+13/51
P(S3)=P(S3/S01*S02)+P(S3/S1*S02)+P(S3/S1*S2)=13/50+12/50+11/50

2. ## Re: Conditional Probability

I think the problem is more complicated than that. Your answer should be a probability distribution like this -
P(0 spades after 3 cards) = ...
P(1 spades after 3 cards) = ...
P(2 spades after 3 cards) = ...
P(3 spades after 3 cards) = ...
Here is one way of attacking the problem.
Let's just look at P(2 spades after 3 cards) = ... The possibilities are SSO, SOS, and OSS. These need to be calculated and added.
Just looking at one of them SOS. The probability is 13/52x39/51x12/50.
You can also draw a probability tree.

3. ## Re: Conditional Probability

You can also solve this problem using the hypergeometric distribution.

For example, P(1S in 3 draws)= [(13C1)(39C2)]/(52C3)

4. ## The Following User Says Thank You to Buckeye For This Useful Post:

Roohul Amin (03-29-2017)

5. ## Re: Conditional Probability

Thanks Buckeye, your approach fit will the situation, that is the condition of without replacement.

6. ## Re: Conditional Probability

Of course. The old hypergeometric trick.

7. ## Re: Conditional Probability

Do you not agree upon the Old Hypergeometric trick? if there is anything wrong with it, please tell.

8. ## Re: Conditional Probability

No. it's perfect. (I was just quoting Maxwell Smart.) I wish I had thought of it before I went to first principles but it gives the same answers.

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