Given that a throw of three unbiased dice shows different faces, what is the probability that their total is…
Given that a throw of three unbiased dice shows different faces, what is the probability that their total is eight?
$\frac{1}{10}$
$\frac{23}{256}$
$\frac{13}{36}$
$\frac{17}{20}$
Solution
Number of ways to get different faces on the a throw of three unbiased dice are ${ }^6 C_3=\frac{6 \times 5 \times 4}{3 \times 2}=20$ and the number of ways to get their total is eight $(1,2,5),(1,3,4)$ only. So, the required probability $=\frac{2}{20}=\frac{1}{10}$ Hence, option (1) is correct.