# A dice is thrown twice. A success is an even number on each throw. Find the probability distribution of the number of successes.

Last updated date: 16th Mar 2023

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Answer

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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.

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