# Help me please: difficult problems

• October 26th 2008, 11:11 AM
bondvista
Hi everyone
I got an exam tomorrow and I just found this website I shoulda found earlier. Please help me if you know how to do these problems:

1. If 3 people are picked from a group of 4 married couples, what is the probability of NOT including a pair of spouses?

2. Find a formula for the probability that EXACTLY one of the events A, B, or C occurs. Then, make up a problem and solve the problem using the formula.

3. Find the probability of getting three of a kind in a regular poker (excluding 2 jokers so there is 52 pokers)

a) 3 of a kind means KKK A 2
My answer is [(13C1)*(4C3) + 2*[(12C1)*(4C1)]] / (52C5)

b) full house: [(13C1)*(4C3) + (12C1)*(4C2)] / (52C5)

c) 4 of a kind: [(13C1)*(4C4) + (12C1)*(4C1)] / (52C5)

What do you guys think?
• October 26th 2008, 02:46 PM
awkward
Quote:

Originally Posted by bondvista
Hi everyone
I got an exam tomorrow and I just found this website I shoulda found earlier. Please help me if you know how to do these problems:

1. If 3 people are picked from a group of 4 married couples, what is the probability of NOT including a pair of spouses?

2. Find a formula for the probability that EXACTLY one of the events A, B, or C occurs. Then, make up a problem and solve the problem using the formula.

3. Find the probability of getting three of a kind in a regular poker (excluding 2 jokers so there is 52 pokers)

a) 3 of a kind means KKK A 2
My answer is [(13C1)*(4C3) + 2*[(12C1)*(4C1)]] / (52C5)

b) full house: [(13C1)*(4C3) + (12C1)*(4C2)] / (52C5)

c) 4 of a kind: [(13C1)*(4C4) + (12C1)*(4C1)] / (52C5)

What do you guys think?

Hi bondvista,

3 of a kind:
$\frac{\binom{13}{1} \binom{4}{3} \binom{12}{2} \binom{4}{1}^2} {\binom{52}{5}}$

Full house:
$\frac{\binom{13}{1} \binom{4}{3} \binom{12}{1} \binom{4}{2}} {\binom{52}{5}}$

(You seem to be adding in some cases above where you should be multiplying.)
• October 26th 2008, 03:01 PM
bondvista
• October 26th 2008, 03:22 PM
Plato
Quote:

Originally Posted by bondvista

On #1. There are 24 ways to choose a married couple to be among the three chosen.
So take that away from the total: ${8 \choose 3}$.

For #2. $P(A \cap B^c \cap C^c ) + P(A^c \cap B \cap C^c ) + P(A^c \cap B^c \cap C)$
• October 26th 2008, 04:08 PM
bondvista
Quote:

Originally Posted by Plato
On #1. There are 24 ways to choose a married couple to be among the three chosen.
So take that away from the total: ${8 \choose 3}$.

For #2. $P(A \cap B^c \cap C^c ) + P(A^c \cap B \cap C^c ) + P(A^c \cap B^c \cap C)$

24 ways. Is this how u got it, 4C1*3C1*2=24?
Thank you guys so much!