A) 4
B) 3
C) 5
D) 8
Correct Answer: B
Solution :
| let \[S\]be the sample space, then \[n\left( S \right)\text{ }=\]total number of determinants that can be made with 0 and 1=\[2\times 2\times 2\times 2=16\] |
| \[\because \left| \begin{align} & a\,\,\,b \\ & c\,\,\,d \\ \end{align} \right|\], each element can be replaced by two types i.e., 0 and 1 |
| And let E be the event that the determinant made is non-negative. |
| Also, E? be the event that the determinant is negative. |
| \[\therefore \,E'=\left\{ \left| \begin{align} & 1\,\,\,1 \\ & 1\,\,\,0 \\ \end{align} \right|,\left| \begin{align} & 0\,\,\,1 \\ & 1\,\,\,\,1 \\ \end{align} \right|,\left| \begin{align} & 0\,\,\,\,1 \\ & 1\,\,\,\,\,0 \\ \end{align} \right| \right\}\] |
| \[\therefore P\left( E' \right)=3\] |
| Then P (E?)\[=\frac{n(E')}{n(S)}=\frac{3}{16}\] |
| Hence, the required probability, |
| \[P(E)=1-P(E')=1-\frac{3}{16}=\frac{13}{16}=\frac{m}{n}\left[ given \right]\] |
| \[\Rightarrow m=13\]and n=16, then n-m=3 |
You need to login to perform this action.
You will be redirected in
3 sec
