Answer
Verified
423.6k+ views
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.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Using the following information to help you answer class 12 chemistry CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Write an application to the principal requesting five class 10 english CBSE
Difference Between Plant Cell and Animal Cell
a Tabulate the differences in the characteristics of class 12 chemistry CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
Discuss what these phrases mean to you A a yellow wood class 9 english CBSE
List some examples of Rabi and Kharif crops class 8 biology CBSE