Seven fair dice are tossed independently. Find the probability that the number of 1's minus the number of 4's will be three.
Thoughts?
What are all of the possible outcomes that satisfy that condition?
# of 1's # of 4's # of (not 1's, not 4's) 3 0 4 4 1 2 5 2 0
Those are the only outcomes that you care about. So, what are the probabilities of each?
I will help you out for the first case, see if you can figure out the other two. The probability that three out of the seven dice will roll a 1 and the other four will roll neither a 1 nor a 4: $\displaystyle \binom{7}{3}\left(\dfrac{1}{6}\right)^3\left( \dfrac{4}{6} \right)^4$
So is it just: $\displaystyle P = \binom{7}{3}\left(\dfrac{1}{6}\right)^3\left( \dfrac{4}{6} \right)^4 + \binom{7}{4}\left(\dfrac{1}{6}\right)^4\binom{7}{1 }\left(\dfrac{1}{6}\right)\left( \dfrac{4}{6} \right)^2 + \binom{7}{5}\left(\dfrac{1}{6}\right)^5\binom{7}{2 }\left(\dfrac{1}{6}\right)^2$?
Not quite. You have $\displaystyle \binom{7}{4}\left(\dfrac{1}{6}\right)^4\binom{7}{1 }\left(\dfrac{1}{6}\right)\left( \dfrac{4}{6} \right)^2$
This means out of seven dice, you choose 4 to roll 1's, then you replace those four dice and choose 1 to roll a 4. You don't want to replace the dice. So, it would be:
$\displaystyle P = \binom{7}{3}\left(\dfrac{1}{6}\right)^3\left( \dfrac{4}{6} \right)^4 + \binom{7}{4}\left(\dfrac{1}{6}\right)^4\binom{3}{1 }\left(\dfrac{1}{6}\right)\left( \dfrac{4}{6} \right)^2 + \binom{7}{5}\left(\dfrac{1}{6}\right)^5\binom{2}{2 }\left(\dfrac{1}{6}\right)^2$