JEE Main 2022 — Probability Question with Solution

Total number of numbers from given

Condition = n(s) = 2⁶.

Every required number is of the form

A = 7 . (10^a₁ + 10^a₂ + 10^a₃ + .......) + 111111

Here 111111 is always divisible by 21.

$$\therefore$$ If A is divisible by 21 then

10^a₁ + 10^a₂ + 10^a₃ + ....... must be divisible by 3.

For this we have ⁶C₀ + ⁶C₃ + ⁶C₆ cases are there

$$\therefore$$ n(E) = ⁶C₀ + ⁶C₃ + ⁶C₆ = 22

$$\therefore$$ Required probability = $$\frac{22}{2^6}=p$$

$$\therefore$$ $$\frac{11}{32}=p$$

$$\therefore$$ $$96p=33$$