Question

A die is thrown twice. What is the probability that $5$ will not come up either time?

Answer 1

Probability of an event $=\frac{numberoffavorableoutcomes}{totalnumberofoutcomes}$

We know that,

If a die is rolled twice, the total number of outcomes is $36$.

Sample space is defined by $\left(\mathrm{S}\right)$

$$\begin{gathered} \mathrm{S}=\mathrm{\{}(1,1)(1,2)(1,3)(1,4)(1,5)(1,6) \\ (2,1)(2,2)(2,3)(2,4)(2,5)(2,6) \\ (3,1)(3,2)(3,3)(3,4)(3,5)(3,6) \\ (4,1)(4,2)(4,3)(4,4)(4,5)(4,6) \\ (5,1)(5,2)(5,3)(5,4)(5,5)(5,6) \\ (6,1)(6,2)(6,3)(6,4)(6,5)(6,6)\} \end{gathered}$$

So, $\mathrm{n}\left(\mathrm{S}\right)=36$

Step 1: Calculating the probability of getting $5$ on either time

Let $\mathrm{A}$ be the event of getting $5$ either time.

The outcomes where $5$ turns up ether time are $=\left\{\right(1,5),(2,5),(3,5),(4,5),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,5\left)\right\}$

So, $\mathrm{n}\left(\mathrm{A}\right)=11$

Therefore, the probability of getting $5$ on either time, $\mathrm{P}\left(\mathrm{A}\right)=\frac{\mathrm{n}\left(\mathrm{A}\right)}{\mathrm{n}\left(\mathrm{S}\right)}$

$\mathrm{P}\left(\mathrm{A}\right)=\frac{11}{36}$

Step 2: Calculating the probability of not getting $5$ on either time

So, the probability that $5$ will not come up on either time $=1–\mathrm{P}\left(\mathrm{A}\right)$

$$\begin{gathered} =1–\frac{11}{36} \\ =\frac{25}{36} \end{gathered}$$

Hence, the probability that $5$ will not come up either time is $\frac{25}{36}$.