Question

If you roll 5 dice what is the probability of getting exactly 3 dices with 4 and two something other than 4

Answer 1

When you roll a die, you get 4 with probabiliy 1/6 and something other than 4 with probability 5/6.

You want to get 4 three times, this happens with probability (1/6)^3.

You want to get something other than 4 two times, this happens with probability (5/6)^2.

Finally, we have to count all the combinations of three rolls ending in 4 and two rolls ending in something other than 4. For example, you might have

44NN4

4N4N4

N44N4

(where '4' represents a roll ending in 4 and 'N' represents a roll ending in something other than 4)

The number of different combinations is given by the binominal coefficient (5 3), because choosing the roll that must end in 4 automatically fixes also those that won't end in 4:

$\displaystyle \binom{5}{3}=\dfrac{5!}{3!2!}=\dfrac{5\times 4\times 3\times 2}{3\times 2\times 2}=10$

So, the final anwer is

$10\left(\dfrac{1}{6}\right)^3\cdot\left(\dfrac{5}{6}\right)^2$