There are five red chips and three blue chips in a bowl. The red chips are numbered 1, 2, 3, 4, 5, respectively, and the blue chips are numbered 1, 2, 3, respectively. If two chips are to be drawn at random and without replacement, find the probability that these chips have either the same number or the same color.
Here is what I am thinking but I don't know if it is correct.
Same number: If you pick one out of six chips (the chips that have other chips with their numbers), then for the second chip, there is 1 out of 5 chips that have the same number. Or you can pick one out of the two that don't have chips with the same number as others. So I think it is:
${6 \choose 2}+{2 \choose 2}$
or maybe it is multiplied and not added together.
Same Color:
${5 \choose 2}+ {3 \choose 2}$
The total number of outcomes is:
${8 \choose 2}$
So that is the denominator in the probability fraction.
So I am thinking:
$\left(\frac{(same \ color) \ + \ (same \ number)}{8 \choose 2}\right)$
$=\left(\frac{{6 \choose 2} \ + \ {2 \choose 2} \ + \ {5 \choose 2} \ + \ {3 \choose 2}}{8 \choose 2}\right) $
That was my thought process, that like I said can be totally wrong.
Can someone please walk me through the problem?
Best Answer
Denominator is indeed $8 \choose 2$
For the numerator:
Same number: There are just 3 options! Both 1's, both 2's, or both 3's. When you multiply $6\choose 1$ and $5 \choose 1$ you are effectively saying that you can pick 1 of the 6 that have a 'mate', and then you can pick any of the 5 others ... that is of course not how it works. There is only 1 option left at that point ... and since order does not matter, you need to further divide by 2, which is how you would get at 3. But common sense tells you that there are just 3 possibilities here like I said: both 1's, both 2's, or both 3's.
Same color: Both red: ${5 \choose 2}= 10$ options (instead of 20 .. order does not matter!) Both blue ${3 \choose 2}= 3$ options (again, order does not matter)