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?

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- Oct 16th 2013, 06:02 PMvidomagruFair Dice Tossed Independently
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? - Oct 16th 2013, 06:27 PMSlipEternalRe: Fair Dice Tossed Independently
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: $\displaystyle \binom{7}{3}\left(\dfrac{1}{6}\right)^3\left( \dfrac{4}{6} \right)^4$ - Oct 16th 2013, 06:50 PMvidomagruRe: Fair Dice Tossed Independently

So is it just: $\displaystyle P = \binom{7}{3}\left(\dfrac{1}{6}\right)^3\left( \dfrac{4}{6} \right)^4 + \binom{7}{4}\left(\dfrac{1}{6}\right)^4\binom{7}{1 }\left(\dfrac{1}{6}\right)\left( \dfrac{4}{6} \right)^2 + \binom{7}{5}\left(\dfrac{1}{6}\right)^5\binom{7}{2 }\left(\dfrac{1}{6}\right)^2$? - Oct 16th 2013, 06:53 PMSlipEternalRe: Fair Dice Tossed Independently
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:

$\displaystyle P = \binom{7}{3}\left(\dfrac{1}{6}\right)^3\left( \dfrac{4}{6} \right)^4 + \binom{7}{4}\left(\dfrac{1}{6}\right)^4\binom{3}{1 }\left(\dfrac{1}{6}\right)\left( \dfrac{4}{6} \right)^2 + \binom{7}{5}\left(\dfrac{1}{6}\right)^5\binom{2}{2 }\left(\dfrac{1}{6}\right)^2$