# Two dice are thrown together. The probability of getting the same number on both the dice is:

$

A.{\text{ }}\dfrac{1}{2} \\

B.{\text{ }}\dfrac{1}{3} \\

C.{\text{ }}\dfrac{1}{6} \\

D.{\text{ }}\dfrac{1}{{12}} \\

$

Last updated date: 23rd Mar 2023

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Answer

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Hint- Solve the problem by the help of sample space of the event and basic definition of probability.

When two dice are thrown together, the total number of all possible outcomes equals to

${{\text{6}}^2} = 36$

This can also be shown through the sample space.

The favorable outcome of getting the same number on both die is

$\left( {1,1} \right),\left( {2,2} \right),\left( {3,3} \right),\left( {4,4} \right),\left( {5,5} \right),\left( {6,6} \right)$

Therefore the number of favorable outcomes is 6

As we know that probability of any event is equal to total number of favorable outcomes upon total number of outcomes.

Hence required probability${\text{ = }}\dfrac{6}{{36}} = \dfrac{1}{6}$

So, option$C$is the correct option.

Note- To find the probability of any event, always calculate the number of favorable cases i.e. the required ones and divide it by the total number of outcomes. In some basic problems try to solve by using sample space of the event.

When two dice are thrown together, the total number of all possible outcomes equals to

${{\text{6}}^2} = 36$

This can also be shown through the sample space.

The favorable outcome of getting the same number on both die is

$\left( {1,1} \right),\left( {2,2} \right),\left( {3,3} \right),\left( {4,4} \right),\left( {5,5} \right),\left( {6,6} \right)$

Therefore the number of favorable outcomes is 6

As we know that probability of any event is equal to total number of favorable outcomes upon total number of outcomes.

Hence required probability${\text{ = }}\dfrac{6}{{36}} = \dfrac{1}{6}$

So, option$C$is the correct option.

Note- To find the probability of any event, always calculate the number of favorable cases i.e. the required ones and divide it by the total number of outcomes. In some basic problems try to solve by using sample space of the event.

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