Question

Each of the faces of a fair six-sided number cube is numbered with one of the numbers $1$ through $6$, with a different number appearing on each face.

Two such number cubes will be tossed, and the sum of the numbers appearing on the faces that land up will be recorded.

What is the probability that the sum will be $4$, given that the sum is less than or equal to $6$ ?

Answer 1

Explanation to correct answer:

We know that

$\mathrm{P}\left(\mathrm{event}\right)=\frac{\mathrm{No}.\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{outcomes}}$

Total outcomes$$\begin{gathered} =(1,1),(1,2),(1,3),(1,4),(1,5),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(4,1),(4,2),(5,1) \\ =15 \end{gathered}$$

Number of outcomes that have a sum equal to $4=(1,3),(2,2),(3,1)$

$\begin{array}{rcl}\therefore P\left(event\right)& =& \frac{3}{15}\\ & =& \frac{1}{5}\end{array}$

Hence, the probability that the sum of the numbers appearing on the faces that land up will be four is $\frac{1}{5}$.