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
Verified
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: $P\left( {A{\text{ or }}B} \right) = P\left( A \right) + P\left( B \right)$.
×
Sorry!, This page is not available for now to bookmark.