Mathematics - Probability Question with Solution | TestHub

For sum to be seven, we need seven$1\mathrm{s}$and two$0s$
$$Number of different possible matrices=$=\frac{\overline{)9}}{\overline{)7}\overline{)2}}=36$
Eg.$\left[\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 0& 0& 1\end{array}\right]${arrangement of seven$1s$and two$0s$}
Now, for determined of matrix to be zero, both the zero must be in same row or in same column.
Eg.$\left[\begin{array}{ccc}1& 0& 0\\ 1& 1& 1\\ 1& 1& 1\end{array}\right]$Now, arrangement of 0s and$1s=\underset{\begin{array}{c}Selectionof\\ a\mathrm{c}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{n}/\mathrm{r}\mathrm{o}\mathrm{w}\end{array}}{6}\times \underset{\begin{array}{c}arrangement\\ oftwo0s\\ andone1in\\ a\mathrm{r}\mathrm{o}\mathrm{w}/\mathrm{c}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{n}\end{array}}{3}=18$
Hence, required probability$=\frac{18}{36}=\frac{1}{2}$
(Note: If two zero's lies on different row/column then$\left|\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 1\end{array}\right|=+1\ne 0$)