# Married couple raffle question.

• Mar 7th 2010, 06:49 PM
Kakariki
Married couple raffle question.
Hey, I don't think I am doing this question correctly, my solutions just seem to simple.

Question:
There are 8 opposite-sex married couples at a party. Two people are chosen at random to win a door prize.
a) What is the probability that the 2 people will be married to each other?
b) What is the probability that the 2 people will be of the same sex?
c) If 6 people are chosen, what is the probability that they are 3 married couples?

Solution:
a) $n(s) = 8C2 = 120$
$n(married couples) = 8$
So, that means that it's 8/120, or 1/15.

b) $(\frac {8}{16})(\frac {7}{15})$
$= \frac {7}{30}$

c) $n(s) = 16C6 = 8008$
There are 8 married couples, so 8C3.
$\frac {56}{8008}$
$\frac {1}{143}$

I am unsure if any of these are correct, as I am not used to doing these sorts of problems with things like married couples.
• Mar 7th 2010, 08:41 PM
Soroban
Hello, Kakariki!

The problems are simple, but require some careful thinking.
And you did fine!

Quote:

There are 8 opposite-sex married couples at a party.
Two people are chosen at random to win a door prize.

a) What is the probability that the 2 people will be married to each other?

Solution:

$n(s) \:=\: _8C_2 \:=\: 120$

$n(\text{married couples}) \:=\: 8$

So, that means that it's: . $\frac{8}{120} \:=\:\frac{1}{15}$ . . Right!

Quote:

b) What is the probability that the 2 people will be of the same sex?
The first person can be any of the 16 people: . $\frac{16}{16} \:=\:1$

The second must be one of the other 7 people of the same sex: . $\frac{7}{15}$

Therefore: . $P(\text{same sex}) \;=\;\frac{7}{15}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Another approach . . .

There are 120 possible outcomes.

$\text{We want: }\:\begin{Bmatrix} \text{2 men:} & _8C_2 \:=\:28\text{ ways} \\

\text{or} \\ \text{2 women: } & _8C_2\:=\:28\text{ ways} \end{Bmatrix}$

Therefore: . $P(\text{same sex}) \;=\;\frac{56}{120} \;=\;\frac{7}{15}$

Quote:

c) If 6 people are chosen, what is the probability that they are 3 married couples?

$n(s) \:=\:_{16}C_6 \:=\:8008$

There are 8 married couples, so: . $_8C_3 \:=\:56$

Therefore: . $\frac{56}{8008} \:=\:\frac{1}{143}$ . . Yes!