MHB What Is the Probability Distribution for Drawing Spades Without Replacement?

[Roohul Amin]

Dear All sorry for repeated post;
There is a problem
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

Need your opinion Please!

 

[Jameson]

Hi there,

So a probability distribution is a function of some kind of input parameter(s). Are you asking what is the probability of drawing three spades or trying to generalize the distribution in terms of the three draws?

In either case the final result would be a multiplication, not a sum. I think you are in the right direction but some clarification on the question would be helpful.

If you want an exhaustive list of possible outcomes, the event is spade/no spade, so the total number of outcomes is $2^3=8$.

 

[Roohul Amin]

Thanks for your reply Jamson: it seems the hyper-geometric distribution fit well this situation, what do you think?

 

[Opalg]

Dear All sorry for repeated post;
There is a problem
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

Need your opinion Please!

I am not entirely sure that I understand the question. But I think that it is asking for the probabilities for drawing 0, 1, 2 or 3 spades.

Let $P(n)$ denote the probability of drawing $n$ spades. Then $P(0) = \dfrac{39 \cdot 38 \cdot 37}{52 \cdot 51 \cdot 50} \approx 0.4135$.

For $P(1)$ we have to draw one spade and two non-spades. There are three possibilities, since the one spade could be either the first, the second or the third card drawn. So $P(1) = \dfrac{3 \cdot 39 \cdot 38 \cdot 13}{52 \cdot 51 \cdot 50} \approx 0.4359.$

In a similar way, you can calculate $P(2)$ and $P(3)$.

If I am interpreting the question correctly, you answer should be the set of four probabilities $P(0),\ldots,P(3)$. As a check, you should verify that your four probabilities add up to $1$.

 

[Roohul Amin]

in the denominator the values are clearly 52C3. But the values in numerator i.e 3.52.52.39 are making some confusion to me, would you pleas give details. Thanks

 

[Opalg]

in the denominator the values are clearly 52C3. But the values in numerator i.e 3.52.52.39 are making some confusion to me, would you pleas give details. Thanks

Sorry, my mistake. I should have written 3.39.38.13 there (I have now corrected it). The idea is that if you want just one of the three cards to be a spade, then you have 39 choices for the first non-spade, 38 for the second non-spade, and 13 for the spade. The 3 (as I mentioned in my previous comment) comes from the fact that the spade could be either the first, the second, or the third card to be chosen.