Q256 / 270 · NDA 2022 (Phase 2) · Mathematics
Three fair dice are tossed once. What is the probability that they show different numbers that are in $A P$ ?
- A$\frac{1}{12}$
- Correct answerB$\frac{1}{18}$
- C$\frac{1}{36}$
- D$\frac{1}{72}$
Solution
$n(s)=6 \times 6 \times 6$
Different numbers that are in AP.
$\begin{aligned}
& (1,2,3),(2,3,4),(3,4,5),(4,5,6),(1,3,5),(2,4, \\
& 6),(3,2,1),(4,3,2),(5,4,3),(6,5,4),(5,3,1),(6, \\
& 4,2) \\
& n(E)=12 \\
& P(E)=\frac{12}{6 \times 6 \times 6}=\frac{1}{18}
\end{aligned}$
Hence, the correct answer is option: 'B'.