1 - Pr(void in at least one suit).

To calculate Pr(void in at least one suit), first read this thread: http://www.mathhelpforum.com/math-he...obability.html

Now modify the argument for five cards rather than 13 cards.

Results 1 to 3 of 3

- September 6th 2008, 09:45 PM #1

- Joined
- Aug 2008
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- 12

## Me again

What is the probability that a five-card hand is not void in any suit?

Answer: .2637

2 spades, 2 hearts, 1 diamond + 3 spades, hearts + all spades?

Im at a loss here

Four men and five women line up so that there is at most one pair of men standing next too each other?

answer = 126,000

5! women being anywhere

4 choose 2 for how many different pairs of men there can be.

4 possible spots for the pair of men to go?

- September 6th 2008, 09:55 PM #2
1 - Pr(void in at least one suit).

To calculate Pr(void in at least one suit), first read this thread: http://www.mathhelpforum.com/math-he...obability.html

Now modify the argument for five cards rather than 13 cards.

- September 7th 2008, 08:30 PM #3

- Joined
- Aug 2008
- Posts
- 12

## k

I looked over it and know that 52 choose 5 is the amount of possible hands.

4(39 choose 5) would be the number to void one card.

What is the probability of a 5 card hand**not**being void in any suit?

So I was looking to subtract... and couldnt find a way to get .2637, I got .22 sumthing