JEE Main 2020 — Probability Question with Solution

Total number of five-digit numbers are $9\left({10}^4\right)$.

Case I : Choosing two non-zero digits i.e. in ${}^{9}C_2$ ways.

So, number of five-digit numbers containing both digits $=2^5-2$.

Case II : Choose one non-zero and one zero as digit i.e. in ${}^{9}C_1$ ways.

So, number of five-digit numbers containing one non-zero and one zero both $=2^4-1$.

So, favourable number of ways $={}^{9}C_2\left(2^5-2\right)+{}^{9}C_1\left(2^4-1\right)=36\left(30\right)+9\left(15\right)=1080+135=1215$

Thus, required probability $=\frac{1215}{9\times {10}^4}=\frac{135}{{10}^4}$.