Question

The corners of regular tetrahedrons are numbers $1,2,3,4$. Three tetrahedrons are tossed. The probability that the sum of upward corners will be $5$ is

• A. $\frac{5}{24}$
• B. $\frac{5}{64}$
• C. $\frac{3}{32}$
• D. $\frac{3}{16}$

Answer 1

The correct option is C

$\frac{3}{32}$

Explanation for correct option:

We know that probability of an event $E$ is defined by $P\left(E\right)=\frac{n\left(E\right)}{n\left(s\right)}$, where $n\left(E\right),n\left(s\right)$ are number of favorable outcomes and number of possible outcomes respectively.

Given that here are $3$ tetrahedrons and each tetrahedron has $4$ corners

So the total number of outcomes is $n\left(s\right)=4^3=64$

The favorable cases that the sum of upward corners will be $5$ $=(2,2,1),(1,2,2),(2,1,2),(1,3,1,),(3,1,1)$ and $(1,1,3)$

So total number of favorable cases $n\left(E\right)=6$

The required probability $P\left(E\right)=\frac{6}{64}$

$=\frac{3}{32}$

Hence option(C) is correct.