More probability, help with cards.

1. May 12, 2012

charmedbeauty

1. The problem statement, all variables and given/known data

What is the probability that a hand of 8 cards dealt from a shuffled pack contains:

b)exactly three cards in at least one of the suits.

2. Relevant equations

3. The attempt at a solution

So

1)choose a suit (4 ways)
2)choose 3 cards from the 13 cards in the suit (C(13,3))
3) choose remaining 5 cards from the remaining 3 suits. (C(39,5)).

so we have 4C(13,3)C(39,5)

Hence the probability is (4C(13,3)C(39,5))/C(52,8)

where C(52,8) is the range of the possible set.

But the top line is wrong, why?

2. May 12, 2012

tiny-tim

hi charmedbeauty!
you're double-counting …

if in 1)and 2) you choose 3 hearts, in 3) you could choose 3 diamonds (+ 1 club + 1 spade)

if in 1)and 2) you choose 3 diamonds, in 3) you could choose 3 hearts (+ 1 club + 1 spade) …

same thing!

3. May 12, 2012

charmedbeauty

Thanks Tiny Tim for the help...
I realised that and figured out the sltn using inclusion/exclusion with 4 subsets...
but is there any easier way to do this??
I figured it out but I knew the answer, but it seemed a very difficult(for me) sltn... is there any easier way?

also for part c) exactly three cards in exactly one of the suits.

Should I just use inclusion /exclusion again??

4. May 12, 2012

tiny-tim

(4? 8/3 < 3, so don't you only need to do it once?)

sorry, but there's usually no quick way to do these questions
nooo, there is a quicker way

(what do the other cards have to be?)

5. May 12, 2012

charmedbeauty

not of the same suit

so initially choose a suit 4C(13,3) with 3 cards from selected suit

then fill rest of hand

C(39,5) from the remaining suits

but now I have overcounted again

so -3C(13,5)

take away the rest of the possibilities of choosing 5 cards from 4 suits, ie we could only have one suit so take away the other three.

better?

6. May 12, 2012

tiny-tim

if only one suit has 3 cards, the distribution must be 3,2,2,1

7. May 12, 2012

charmedbeauty

I don't follow
the distribution of what? suits?

I only one suit has three cards

there are 5 places remaining to fill with 3 suits.

so it could be 5 from one suit

4 from one suit and 1 from another

2 from one suit, 2 from the other, and one from the last suit

and they are the only combinations possible since we cant introduce another three cards from the same suit.

I dont see why I should only choose the intersection of the remaining suits.?

why cant I have my 8 card hand consisting of 3 from one suit, and 5 from another?

8. May 12, 2012

tiny-tim

oh i misunderstood

i thought 3 was the maximum

9. May 12, 2012

charmedbeauty

ok so is this sltn correct...

4C(13,3)C(39,5)-3C(13,3)

??

10. May 12, 2012

tiny-tim

no, you need to subtract the number of ways of getting exactly 3 in 2 suits

11. May 12, 2012

Ray Vickson

You could count like this: you need either (1) three cards from just one of the suits; and (2) three cards from each of two suits.

In (1): if, for example, your three cards are hearts, you need the other five cards from spades, clubs and diamonds, choosing no more than two from each of these. That means that you do, in fact, choose two from each of two suits and one from the other suit. How many ways are there to do that? There are C(13,3) ways of choosing three hearts. There are C(3,2) = 3 ways of choosing the two suits that have doubles, and for each such way there are C(13,2)^2 ways of choosing the doubles; then there are 13 ways of choosing the remaining card. Now, of course, the three cards are not hearts, you need to do the same calculations for each of the other choices. Altogether, the number of ways in case (1) is 4*3*13*C(13,3)*C(13,2)^2.

You can do a similar calculation in case (2). Adding them up gives you the numerator.

RGV

Last edited: May 12, 2012
12. May 13, 2012

charmedbeauty

ok so i need to subtract -3C(13,3)C(26,3)?

13. May 13, 2012

charmedbeauty

ok so i need to subtract -3C(13,3)C(26,3)?

14. May 13, 2012

tiny-tim

(just got up :zzz:)
(did you mean 3C(13,3)C(26,2)? anyway …)

that only gives you 6 (or 5) cards

15. May 13, 2012

charmedbeauty

I thought it should be C(26,3) Since you said "no, you need to subtract the number of ways of getting exactly 3 in 2 suits".

isn't that C(26,3)??

16. May 14, 2012

tiny-tim

oh i see!

no i meant eg exactly three hearts and exactly three diamonds

17. May 14, 2012

charmedbeauty

oh ok, so its C(25,2).

thanks for the help tinytim