# 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

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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 $\dfrac{1}{6}$. 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 $\dfrac{1}{6}$.

So, the probability of not getting $5$ on first rolling is $1 - \dfrac{1}{6}$ which is $\dfrac{5}{6}$.

Similarly, the probability of not getting $5$ on second rolling is also $\dfrac{5}{6}$.

And since both the outcomes are independent, the probability of not getting $5$ on either of the time is:

$P = \dfrac{5}{6} \times \dfrac{5}{6} = \dfrac{{25}}{{36}}$.

Hence, the required probability is$\dfrac{{25}}{{36}}$.

$(ii)$ 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:

$P = \dfrac{1}{6} \times \dfrac{5}{6} = \dfrac{5}{{36}}$.

In the second case, we get any other number the first time and $5$ second time. Probability in this case will be:

$P = \dfrac{5}{6} \times \dfrac{1}{6} = \dfrac{5}{{36}}$.

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:

$

\Rightarrow P = \dfrac{5}{{36}} + \dfrac{5}{{36}}, \\

\Rightarrow P = \dfrac{{10}}{{36}}, \\

\Rightarrow P = \dfrac{5}{{18}}. \\

$

Hence, the required probability is$\dfrac{5}{{18}}$.

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:

$P\left( {A{\text{ or }}B} \right) = P\left( A \right) + P\left( B \right)$.

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 $\dfrac{1}{6}$. 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 $\dfrac{1}{6}$.

So, the probability of not getting $5$ on first rolling is $1 - \dfrac{1}{6}$ which is $\dfrac{5}{6}$.

Similarly, the probability of not getting $5$ on second rolling is also $\dfrac{5}{6}$.

And since both the outcomes are independent, the probability of not getting $5$ on either of the time is:

$P = \dfrac{5}{6} \times \dfrac{5}{6} = \dfrac{{25}}{{36}}$.

Hence, the required probability is$\dfrac{{25}}{{36}}$.

$(ii)$ 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:

$P = \dfrac{1}{6} \times \dfrac{5}{6} = \dfrac{5}{{36}}$.

In the second case, we get any other number the first time and $5$ second time. Probability in this case will be:

$P = \dfrac{5}{6} \times \dfrac{1}{6} = \dfrac{5}{{36}}$.

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:

$

\Rightarrow P = \dfrac{5}{{36}} + \dfrac{5}{{36}}, \\

\Rightarrow P = \dfrac{{10}}{{36}}, \\

\Rightarrow P = \dfrac{5}{{18}}. \\

$

Hence, the required probability is$\dfrac{5}{{18}}$.

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:

$P\left( {A{\text{ or }}B} \right) = P\left( A \right) + P\left( B \right)$.

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