Question

A dice is rolled twice. Find the probability that:
$(i)$$5$ will not come up either time $(ii)$ $5$ will come up exactly one time.

Answer 1

Hint: Both the outcomes of the dice will be independent to each other. Apply the theorem of probability of the independent events.

Since a dice is always $6$ faced, numbered $1$ to $6$, the probability of getting any number from $1$ to $6$ on its rolling is ${\displaystyle \frac{1}{6}}$. And if it’s rolled twice, both the outcomes will be independent to each other.
$(i)$We have to calculate the probability of not getting $5$ on either of the rolling.
As discussed earlier, the probability of getting $5$ on the first rolling is ${\displaystyle \frac{1}{6}}$.
So, the probability of not getting $5$ on first rolling is $1-{\displaystyle \frac{1}{6}}$ which is ${\displaystyle \frac{5}{6}}$.
Similarly, the probability of not getting $5$ on second rolling is also ${\displaystyle \frac{5}{6}}$.
And since both the outcomes are independent, the probability of not getting $5$ on either of the time is:
$P={\displaystyle \frac{5}{6}}\times {\displaystyle \frac{5}{6}}={\displaystyle \frac{25}{36}}$.
Hence, the required probability is${\displaystyle \frac{25}{36}}$.
$(ii)$ Here we have to calculate the probability of getting $5$ exactly one time. Here we’ll have two cases:
Let’s suppose in the first case, we get $5$ on the first time and any other number on the second time. Then the probability will be:
$P={\displaystyle \frac{1}{6}}\times {\displaystyle \frac{5}{6}}={\displaystyle \frac{5}{36}}$.
In the second case, we get any other number the first time and $5$ second time. Probability in this case will be:
$P={\displaystyle \frac{5}{6}}\times {\displaystyle \frac{1}{6}}={\displaystyle \frac{5}{36}}$.
And both the cases are mutually exclusive. Then the total probability of getting $5$ exactly one time is the addition of probability of both the cases:
$$\begin{gathered} \Rightarrow P={\displaystyle \frac{5}{36}}+{\displaystyle \frac{5}{36}}, \\ \Rightarrow P={\displaystyle \frac{10}{36}}, \\ \Rightarrow P={\displaystyle \frac{5}{18}}. \end{gathered}$$
Hence, the required probability is${\displaystyle \frac{5}{18}}$.

Note: If two events $A$ and $B$ are independent to each other, then the probability of occurrence of both the events is:
$P\left(A\text{ and }B\right)=P\left(A\right)\times P\left(B\right)$
While if two events $A$ and $B$ are mutually exclusive to each other, then the probability of occurrence of any one of them is:
$P\left(A\text{ or }B\right)=P\left(A\right)+P\left(B\right)$.