Question

The chance of throwing an ace is the first only of two successive throws with an ordinary die is

• A. $\frac{1}{36}$
• B. $\frac{5}{36}$
• C. $\frac{25}{36}$
• D. $\frac{1}{6}$

Answer 1

Answer: (B)

$\frac{5}{36}$

Probability of throwing an are $=\frac{1}{6}$,
and of not throwing ace $=\frac{5}{6}$ .
Hence required probability $=\frac{1}{6}\times \frac{5}{6}=\frac{5}{36}.$