JEE Main 2021 — Mathematics Probability & Statistics

Given the probability that the digit $0$0 appear at the even place is $\frac{1}{2},$21​, hence the probability that the digit $1$1 appear at the even place is $1-\frac{1}{2}=\frac{1}{2}.$1−21​=21​.

And the probability that the digit $0$0 appear at the odd place is $\frac{1}{3},$31​, hence the probability that the digit $1$1 appear at the odd place is $1-\frac{1}{3}=\frac{2}{3}.$1−31​=32​.

For the digits $'10'$′10′ and $'01'$′01′ we have the following possibilities

$$\begin{matrix}1 & 0 & 0 & 1 \\ oddplace & evenplace & oddplace & evenplace\end{matrix}$$1oddplace​0evenplace​0oddplace​1evenplace​

Or

$$\begin{matrix}1 & 0 & 0 & 1 \\ evenplace & oddplace & evenplace & oddplace\end{matrix}$$1evenplace​0oddplace​0evenplace​1oddplace​

Thus, the required probability is $(\frac{1}{2}\cdot \frac{1}{3}\cdot \frac{1}{2}\cdot \frac{2}{3})+(\frac{2}{3}\cdot \frac{1}{2}\cdot \frac{1}{3}\cdot \frac{1}{2})$(21​⋅31​⋅21​⋅32​)+(32​⋅21​⋅31​⋅21​)

$$=\frac{1}{18}+\frac{1}{18}=\frac{1}{9}.$$=181​+181​=91​.