More probability, help with cards.

[charmedbeauty]

Homework Statement

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.

Homework Equations

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?

 

[tiny-tim]

hi charmedbeauty!

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.
…
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)).

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! ​

 

[charmedbeauty]

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! ​

Thanks Tiny Tim for the help...
I realized 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??

 

[tiny-tim]

Thanks Tiny Tim for the help...
I realized 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?

(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

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

Should I just use inclusion /exclusion again??

nooo, there is a quicker way

(what do the other cards have to be?)

 

[charmedbeauty]

(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?)

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?

 

[tiny-tim]

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

 

[charmedbeauty]

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

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 can't introduce another three cards from the same suit.

I don't see why I should only choose the intersection of the remaining suits.?

why can't I have my 8 card hand consisting of 3 from one suit, and 5 from another?

 

[tiny-tim]

why can't I have my 8 card hand consisting of 3 from one suit, and 5 from another?

oh i misunderstood

i thought 3 was the maximum

 

[charmedbeauty]

oh i misunderstood

i thought 3 was the maximum

ok so is this sltn correct...

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

??

 

[tiny-tim]

ok so is this sltn correct...

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

??

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

 

[Ray Vickson]

Homework Statement

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.

Homework Equations

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?

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

 

[charmedbeauty]

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

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

 

[charmedbeauty]

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

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

 

[tiny-tim]

(just got up :zzz:)

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

(did you mean 3C(13,3)C(26,2)? anyway …)

that only gives you 6 (or 5) cards

 

[charmedbeauty]

(just got up :zzz:)


(did you mean 3C(13,3)C(26,2)? anyway …)

that only gives you 6 (or 5) cards

Im confused about C(26,2)

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)??

 

[tiny-tim]

oh i see!

no i meant eg exactly three hearts and exactly three diamonds

 

[charmedbeauty]

oh i see!

no i meant eg exactly three hearts and exactly three diamonds

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

thanks for the help tinytim