Question

Let $S$ be the sample space of all $3\times 3$ matrices with entries form the set $\{0,1\}$. Let the events $E_1$ and $E_2$ be given by :

$E_1=\{A\in S:\text{det(A)}=0\}$

$E_2=\{A\in S:$ sum of entries of $A$ is $7\}$

If a matrix is chosen at random from $S$, then the conditional probability $P\left(E_1|E_2\right)$equals.

Answer 1

Finding conditional probability $P\left(E_1|E_2\right)$:

Total number of outcomes in $S$ is $2^9$, as there are $9$ places of entries and at each place only $2$ values are possible.

$P\left(\frac{E_1}{E_2}\right)=P\left(\frac{E_1\cap E_2}{E_2}\right)$

$E_2=$ the sum of entries of $A=7$.

To, get the sum values $7$, only one combination is possible which is seven times One and two times zero.

As we know, ${}^n\mathrm{C}_{\mathrm{r}}=\frac{\mathrm{n}!}{\left(\mathrm{n}-\mathrm{r}\right)!\mathrm{r}!}$

Using the Combination formula, to get $E_2=$ $\frac{9!}{7!2!}\left[\because \frac{n!}{(n-r)!r!},n=9,r=2\right]$

$=36$

Also, $\left|A\right|$to be zero, both zeroes should be in the same row or column. That's the only way to make $\left|A\right|$$=0$.

$\left|A\right|=$$\left|\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 0& 0\end{array}\right|$

So, there are $3$ rows and two columns.

Therefore, $E_1=$$3\times 3\times 2=18$ cases (having both zeroes in same column/row)

Thus, conditional probability $P\left(E_1|E_2\right)$$=\frac{18}{36}=\frac{1}{2}$