A dice is thrown twice. A success is an even number on each throw. Find the probability distribution of the number of successes.
Answer
364.8k+ views
Hint: To find the probability distribution, first we will find the probability and then use the data to find the probability distribution.
So, to find the probability we have to find out the number of favourable conditions and total conditions.
Total number of cases when a dice is thrown twice are:
$S = \left\{ {1,2,3,4,5,6} \right\}$, S being the sample space.
$n(S) = 6$
Now, the number of favourable cases, in this case we need only even number outcomes which are $2,4$ and $6$.
Therefore,
$E = \left\{ {2,4,6} \right\}$
Therefore,
$n\left( E \right) = 3$
Now, the next step is to find the probability, the formula of probability is
$P\left( E \right) = \dfrac{{n\left( E \right)}}{{n\left( S \right)}}$,
If we put the values in the above equation, we get,
$P\left( E \right) = \dfrac{3}{6} = \dfrac{1}{2}$
Now, the next step is to find the probability of failure, which is calculated by$1 - P(success)$, because the sum of success and failure is always equal to $1$ in probability.
Probability of failure $ = P(failure) = 1 - \dfrac{1}{2} = \dfrac{1}{2}$
Therefore,
$n = 2$,
So, for no success (i.e.$x = 0$), we are going to find the probability,
$P\left( {x = 0} \right) = {}^2{C_0}{\left( {\dfrac{1}{2}} \right)^0}{\left( {\dfrac{1}{2}} \right)^{2 - 0}} = {\left( {\dfrac{1}{2}} \right)^2} = \dfrac{1}{4}$
Similarly, for a single case of success (i.e$x = 1$), we are going to find the probability,
$P\left( {x = 1} \right) = {}^2{C_{_1}}{\left( {\dfrac{1}{2}} \right)^1}{\left( {\dfrac{1}{2}} \right)^{2 - 1}} = \dfrac{1}{2}$
Similarly, for double case of success (i.e.$x = 2$), we are going to find the probability,
$P\left( {x = 2} \right) = {}^2{C_2}{\left( {\dfrac{1}{2}} \right)^2}{\left( {\dfrac{1}{2}} \right)^{2 - 2}} = \dfrac{1}{4}$
Next step is to find the probability distribution using the above data,
$x = r$ $0$ $1$ $2$
$P\left( {x = r} \right)$ $\dfrac{1}{4}$ $\dfrac{1}{2}$ $\dfrac{1}{4}$
$\sum {P\left( {x = r} \right)} = \dfrac{1}{4} + \dfrac{1}{2} + \dfrac{1}{4} = 1$
Note: Make sure to take the sample space correctly. Here, first we will find the probability and then use the data to find the probability distribution.
So, to find the probability we have to find out the number of favourable conditions and total conditions.
Total number of cases when a dice is thrown twice are:
$S = \left\{ {1,2,3,4,5,6} \right\}$, S being the sample space.
$n(S) = 6$
Now, the number of favourable cases, in this case we need only even number outcomes which are $2,4$ and $6$.
Therefore,
$E = \left\{ {2,4,6} \right\}$
Therefore,
$n\left( E \right) = 3$
Now, the next step is to find the probability, the formula of probability is
$P\left( E \right) = \dfrac{{n\left( E \right)}}{{n\left( S \right)}}$,
If we put the values in the above equation, we get,
$P\left( E \right) = \dfrac{3}{6} = \dfrac{1}{2}$
Now, the next step is to find the probability of failure, which is calculated by$1 - P(success)$, because the sum of success and failure is always equal to $1$ in probability.
Probability of failure $ = P(failure) = 1 - \dfrac{1}{2} = \dfrac{1}{2}$
Therefore,
$n = 2$,
So, for no success (i.e.$x = 0$), we are going to find the probability,
$P\left( {x = 0} \right) = {}^2{C_0}{\left( {\dfrac{1}{2}} \right)^0}{\left( {\dfrac{1}{2}} \right)^{2 - 0}} = {\left( {\dfrac{1}{2}} \right)^2} = \dfrac{1}{4}$
Similarly, for a single case of success (i.e$x = 1$), we are going to find the probability,
$P\left( {x = 1} \right) = {}^2{C_{_1}}{\left( {\dfrac{1}{2}} \right)^1}{\left( {\dfrac{1}{2}} \right)^{2 - 1}} = \dfrac{1}{2}$
Similarly, for double case of success (i.e.$x = 2$), we are going to find the probability,
$P\left( {x = 2} \right) = {}^2{C_2}{\left( {\dfrac{1}{2}} \right)^2}{\left( {\dfrac{1}{2}} \right)^{2 - 2}} = \dfrac{1}{4}$
Next step is to find the probability distribution using the above data,
$x = r$ $0$ $1$ $2$
$P\left( {x = r} \right)$ $\dfrac{1}{4}$ $\dfrac{1}{2}$ $\dfrac{1}{4}$
$\sum {P\left( {x = r} \right)} = \dfrac{1}{4} + \dfrac{1}{2} + \dfrac{1}{4} = 1$
Note: Make sure to take the sample space correctly. Here, first we will find the probability and then use the data to find the probability distribution.
Last updated date: 26th Sep 2023
•
Total views: 364.8k
•
Views today: 5.64k
Recently Updated Pages
What do you mean by public facilities

Slogan on Noise Pollution

Paragraph on Friendship

Disadvantages of Advertising

Prepare a Pocket Guide on First Aid for your School

What is the Full Form of ILO, UNICEF and UNESCO

Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

What is the past tense of read class 10 english CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

How many millions make a billion class 6 maths CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Number of Prime between 1 to 100 is class 6 maths CBSE

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

One cusec is equal to how many liters class 8 maths CBSE

How many crores make 10 million class 7 maths CBSE
