Five people are selected at random. What is the probability that none of the people in this group were born in the same month?

is it 1-(12/12c5)

please help

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- Dec 9th 2007, 07:06 PMAmyUrgent help on probability
Five people are selected at random. What is the probability that none of the people in this group were born in the same month?

is it 1-(12/12c5)

please help - Dec 9th 2007, 07:14 PMDivideBy0
I got

$\displaystyle \frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8}{12^5}$

It's equivalent to placing the five people in twelve chairs. 12 choices for the first, 11 for the second ... etc.

Divded of course by the number of ways they can sit if they are allowed to sit on top of each other :p - Dec 9th 2007, 07:43 PMAmy
- Dec 9th 2007, 07:54 PMDivideBy0
Ok I'm just using:

$\displaystyle Pr(Favourable \ Outcome) = \frac{number \ of \ ways \ favourable \ outcome \ can \ occur}{total \ number \ of \ outcomes}$

We'll simplify this to

$\displaystyle Pr (X) = \frac{n(X)}{n(\varepsilon)}$

where Pr(X) is the probability the 5 people are born in different months and the set $\displaystyle \varepsilon$ represents all possible outcomes.

Perhaps the chair analogy wasn't a very clear one. Let's go back to the months. We need to find the number of ways five people can be born on different months.

There are 12 different months and 5 people, and this is sampling without replacement, so we use permutation:

$\displaystyle n(X) = P_5^{12}= 95040$

And the total number of possibilities for the months the 5 people are born in are given by $\displaystyle n(\varepsilon)=12 \times 12 \times 12 \times 12 \times 12=12^5$ because each person has 12 months to choose from. It's just the multiplication principle for counting.

So $\displaystyle Pr(X) = \frac{95040}{12^5}=\frac{55}{144}$