# A die is thrown once. What is the probability of getting a number 3 or 4?

A.$\dfrac{1}{3}$

B.$\dfrac{2}{3}$

C.$0$

D.$1$

Last updated date: 21st Mar 2023

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Answer

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Hint: Find the sample space of throwing a die single time. Then find the total number of favourable outcomes and then find the probability.

Complete step-by-step answer:

Given a die is thrown once. We need to find the probability of getting a number 3 or 4.

A normal die has a cuboidal structure.

The die has six faces with number 1 to 6 written on them.

On throwing a die only one face can be seen above.

So, the sample space of throwing a single die once is as follow:

$S:\left\{ {1,2,3,4,5,6} \right\}$

We need to find the probability of getting a number 3 or 4.

Hence our favourable outcome is as follow:

$E:\left\{ {3,4} \right\}$

From above sets it is clear that,

Total number of outcomes$ = n\left( S \right) = 6$

And favourable number of outcomes$ = n\left( E \right) = 2$

Now we can find the probability of the favourable outcome.

Probability of getting the favourable outcome$ = \dfrac{{n\left( E \right)}}{{n\left( S \right)}}$

Using the above values, we get

Probability of getting a number 3 or 4 $ = \dfrac{{n\left( E \right)}}{{n\left( S \right)}} = \dfrac{2}{6} = \dfrac{1}{3}$

Hence the correct option is (A). $\dfrac{1}{3}$.

Note: Classical probability problems like above should be solved using sample space. Permutation and combination can be used in case of large sample spaces. Probability can never be negative and its value lies between 0 and 1 inclusive of both.

Complete step-by-step answer:

Given a die is thrown once. We need to find the probability of getting a number 3 or 4.

A normal die has a cuboidal structure.

The die has six faces with number 1 to 6 written on them.

On throwing a die only one face can be seen above.

So, the sample space of throwing a single die once is as follow:

$S:\left\{ {1,2,3,4,5,6} \right\}$

We need to find the probability of getting a number 3 or 4.

Hence our favourable outcome is as follow:

$E:\left\{ {3,4} \right\}$

From above sets it is clear that,

Total number of outcomes$ = n\left( S \right) = 6$

And favourable number of outcomes$ = n\left( E \right) = 2$

Now we can find the probability of the favourable outcome.

Probability of getting the favourable outcome$ = \dfrac{{n\left( E \right)}}{{n\left( S \right)}}$

Using the above values, we get

Probability of getting a number 3 or 4 $ = \dfrac{{n\left( E \right)}}{{n\left( S \right)}} = \dfrac{2}{6} = \dfrac{1}{3}$

Hence the correct option is (A). $\dfrac{1}{3}$.

Note: Classical probability problems like above should be solved using sample space. Permutation and combination can be used in case of large sample spaces. Probability can never be negative and its value lies between 0 and 1 inclusive of both.

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