JEE Main 2023 — Mathematics Probability & Statistics

Given,

$$S={M=[{a}_{ij}],{a}_{ij}\in {0,1,2},{1\leq i,j\leq 2}}$$S=M=[aij​],aij​∈0,1,2,1≤i,j≤2

So, let $M=[\begin{matrix}a & b \\ c & d\end{matrix}]$M=[ac​bd​] where $a,b,c,d\in {0,1,2}$a,b,c,d∈0,1,2

So, total sample space will be ${3}^{4}=81$34=81

Now we will find $P(\bar{A})$P(Aˉ) or possibilities for $|M|=0$∣M∣=0

So, for $|M|=0$∣M∣=0 cases will be,

1.$[\begin{matrix}1 & 1 \\ 1 & 1\end{matrix}]$[11​11​] or $[\begin{matrix}0 & 0 \\ 0 & 0\end{matrix}]$[00​00​] or $[\begin{matrix}2 & 2 \\ 2 & 2\end{matrix}]\rightarrow$[22​22​]→ Total matrix $=3$=3

1. Two $1's$1′s and Two $0's$0′s $\rightarrow$→ Total matrix $=4$=4

2. Two $2's$2′s and Two $0's$0′s $\rightarrow$→ Total matrix $=4$=4

3. Two $1's$1′s and Two $2's$2′s $\rightarrow$→ Total matrix$=4$=4

4. One $1$1 and three $0's$0′s $\rightarrow$→ Total matrix $=4$=4

5. One $2$2 and Three $0's$0′s $\rightarrow$→ Total matrix $=4$=4

6. One $1$1 and one $2$2 and two $0's$0′s $\rightarrow$→ Total matrix $=8$=8

So, total cases for which $|M|=0$∣M∣=0 is $31$31,

So probability will be $P(\bar{A})=\frac{31}{81}$P(Aˉ)=8131​

Hence, $P(A)=1-\frac{31}{81}=\frac{50}{81}$P(A)=1−8131​=8150​