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.)
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?
Hello, alexgintom!
I assume you area familiar with Combinations.;
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: .$\displaystyle _8C_2 \,=\,28\text{ ways to draw 2 numbers.}$
There are: .$\displaystyle _4C_2 \,=\,6\text{ ways to draw 2 even numbers.}$
Therefore: .$\displaystyle P(\text{both even}) \:=\:\frac{6}{28} \:=\:\frac{3}{14}$