# Another probability!

• Apr 28th 2012, 04:08 PM
alexgintom
Another probability!
Each of the numbers 1-8 is written on a piece of paper and placed in a hat. If two numbers are selected at random, find the probability that both numbers selected are even. (The papers are not replaced.)
• Apr 28th 2012, 07:20 PM
Quant
Re: Another probability!
Well hi there!

Do you know the probability of selecting an even number out of 1 to 8? Obviously it is

p(even number out of 1-8) = ...

Now imagine you already selected an even number. The papers are not replaced, so there are 7 papers left. There are still 4 of them labeled with an uneven number. In addition there are 3 even numbers left, because you have already selected an even number in the first draw. So it is

p(even number out of 7 remaining papers) = 3/7.

p(both selected numbers are even) = p(even number out of 1-8) * p(even number out of 7 remaining papers)

Do you know what p(even number out of 1-8) is equal to?
• Apr 29th 2012, 09:22 AM
Soroban
Re: Another probability!
Hello, alexgintom!

I assume you area familiar with Combinations.;

Quote:

Each of the numbers 1-8 is written on a piece of paper and placed in a hat.
If two numbers are selected at random without replacement,
. . find the probability that both numbers selected are even.

There are 8 numbers: 4 even and 4 odd.

There are: . $_8C_2 \,=\,28\text{ ways to draw 2 numbers.}$

There are: . $_4C_2 \,=\,6\text{ ways to draw 2 even numbers.}$

Therefore: . $P(\text{both even}) \:=\:\frac{6}{28} \:=\:\frac{3}{14}$