Fair Dice Tossed Independently

• Oct 16th 2013, 06:02 PM
vidomagru
Fair 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 PM
SlipEternal
Re: 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: $\binom{7}{3}\left(\dfrac{1}{6}\right)^3\left( \dfrac{4}{6} \right)^4$
• Oct 16th 2013, 06:50 PM
vidomagru
Re: Fair Dice Tossed Independently
Quote:

Originally Posted by SlipEternal
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: $\binom{7}{3}\left(\dfrac{1}{6}\right)^3\left( \dfrac{4}{6} \right)^4$

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

Originally Posted by vidomagru
So is it just: $P = \binom{7}{3}\left(\dfrac{1}{6}\right)^3\left( \dfrac{4}{6} \right)^4 + \binom{7}{4}\left(\dfrac{1}{6}\right)^4\binom{7}{1 }\left(\dfrac{1}{6}\right)\left( \dfrac{4}{6} \right)^2 + \binom{7}{5}\left(\dfrac{1}{6}\right)^5\binom{7}{2 }\left(\dfrac{1}{6}\right)^2$?

Not quite. You have $\binom{7}{4}\left(\dfrac{1}{6}\right)^4\binom{7}{1 }\left(\dfrac{1}{6}\right)\left( \dfrac{4}{6} \right)^2$
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:

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