Three dice are rolled. If the probability of getting different...

Given that three dice are thrown.

The total number of outcomes when three dice are thrown together is ${6}^{3} = 6 \times 6 \times 6$.

The number of outcomes such that all the outcomes are different is $= 6 \times 5 \times 4$.

For ex: If the outcome in dice $1$ is ${}^{"}{6}^{"}$ then the number of outcomes for dice $2$ should be $5 ( 1, 2, 3, 4, 5 )$ which excludes the outcome ${}^{"}{6}^{"}$.

Hence, the required probability is $= \dfrac{6 \times 5 \times 4}{6 \times 6 \times 6}$

$= \dfrac{20}{36} = \dfrac{5}{9} = \dfrac{p}{q}$

$⇒ q - p = 9 - 5 = 4$

Therefore, the required answer is $4$.