Three dice are rolled.

Question

Three dice are rolled. If the probability of getting different numbers on the three dice is $$\frac{p}{q}$$, where $p$ and $q$ are co-prime, then $q-p$ is equal to :

Accepted Answer

Total number of outcomes $=6 	imes 6 	imes 6=216$

Number of outcomes getting different numbers on the three dice are ${ }^6 P_3=\frac{6 !}{3 !}=120$

$	herefore$ Required probability $=\frac{120}{216}=\frac{5}{9}$

$	herefore p=5$ and $q=9$

$	herefore q-p=9-5=4$

Solutions · 1