Question

Let S be the sample space of all 3 X 3 matrices with the entries from the set {0, 1}. Let the events {{E}_{1}}=\left\\{ A\in S:\det A=0 \right\\} and ${{E}_{2}}$ = {$A\in S$: Sum of all the entries of A is 7}. If the matrix is chosen at random from S, then the conditional property $P\left( \left. {{E}_{1}} \right|{{E}_{2}} \right)$ equals.

Answer 1

First of all, we know that a 3 3 matrix has 9 spaces. S is the event set of all 3 3 matrices with the entries from the set {0, 1}. Thus, we will find the number of ways in which 1 and 0 can arrange themselves in 9 spaces, with repetition. Then we will find the number of ways in which the event ${{E}_{2}}$ can occur and number of ways in which event ${{E}_{1}}\cap {{E}_{2}}$ can occur. We can find the probability of any event A by the relation $P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}$, where n(A) number of ways A can occur and n(S) is number elements in sample event set S. Using this relation, we will find the probabilities of the event sets ${{E}_{1}}\cap {{E}_{2}}$ and ${{E}_{2}}$. Once we get the probabilities, we can find the conditional probability given by the relation $P\left( {{E}_{1}}\left| {{E}_{2}} \right. \right)=\dfrac{P\left( {{E}_{1}}\cap {{E}_{2}} \right)}{P\left( {{E}_{2}} \right)}$ .

Complete step-by-step answer :
It is given that S is the event set of all 3 3 matrices with the entries from the set {0, 1}.
Thus, each space in the matrix will have 2 options, either 0 or 1.
Therefore, the number of ways in which 2 items can occupy 9 spaces is given as n(S) = ${{2}^{9}}$.
Now, it is given that ${{E}_{2}}$ is an event such that ${{E}_{2}}$ = {$A\in S$: Sum of all the entries of A is 7}.
This means, 7 out of 9 spaces must be occupied by 1 and the remaining 2 spaces will be occupied by 0.
Therefore, number of ways this is possible is given as n(${{E}_{2}}$) = $^{9}{{C}_{7}}$ = 36.
It is also given that {{E}_{1}}=\left\\{ A\in S:\det A=0 \right\\}.
So, the event set such that both events ${{E}_{1}}$ and ${{E}_{2}}$ are happening simultaneously is given as ${{E}_{1}}\cap {{E}_{2}}$.
${{E}_{1}}\cap {{E}_{2}}$ will be true when the two 0’s are in the same row or same column.
Thus, number of ways two zeros can be in the same row is given by $^{3}{{C}_{1}}{{\times }^{3}}{{C}_{2}}$, i.e. selecting one row out of 3 and selecting 2 spaces out of 3 for 0’s.
In similar way, number of ways two zeros can be in the same column is given as $^{3}{{C}_{1}}{{\times }^{3}}{{C}_{2}}$.
Therefore n(${{E}_{1}}\cap {{E}_{2}}$) = $2{{\times }^{3}}{{C}_{1}}{{\times }^{3}}{{C}_{2}}=18$.
Now, we will find the probability of ${{E}_{2}}$.
$\Rightarrow P\left( {{E}_{2}} \right)=\dfrac{36}{{{2}^{9}}}$
We will also find the probability of ${{E}_{1}}\cap {{E}_{2}}$.
$\Rightarrow P\left( {{E}_{1}}\cap {{E}_{2}} \right)=\dfrac{18}{{{2}^{9}}}$
The conditional probability is given as $P\left( {{E}_{1}}\left| {{E}_{2}} \right. \right)=\dfrac{P\left( {{E}_{1}}\cap {{E}_{2}} \right)}{P\left( {{E}_{2}} \right)}$.
$\begin{aligned} & \Rightarrow P\left( {{E}_{1}}\left| {{E}_{2}} \right. \right)=\dfrac{\dfrac{18}{{{2}^{9}}}}{\dfrac{36}{{{2}^{9}}}} \\\ & \Rightarrow P\left( {{E}_{1}}\left| {{E}_{2}} \right. \right)=\dfrac{18}{36} \\\ & \Rightarrow P\left( {{E}_{1}}\left| {{E}_{2}} \right. \right)=\dfrac{1}{2} \\\ \end{aligned}$

Note : It is not necessary to find the number of elements in the sample space S and nor the individual probabilities. We can directly find the conditional probability by the relation $P\left( {{E}_{1}}\left| {{E}_{2}} \right. \right)=\dfrac{n\left( {{E}_{1}}\cap {{E}_{2}} \right)}{n\left( {{E}_{2}} \right)}$ as the denominator is same, i.e. the number of elements of sample set S and gets divided out.