# 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.

Last updated date: 27th Mar 2023

•

Total views: 309.3k

•

Views today: 2.87k

Answer

Verified

309.3k+ views

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:

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

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE