Let A be the set of all 4-digit natural numbers with exactly one digit equal to 7.

Question

Let $A$ be the set of all 4-digit natural numbers with exactly one digit equal to 7. Find the probability that a randomly chosen element of $A$ leaves remainder 2 when divided by 5.

Accepted Answer

Total 4-digit numbers with exactly one digit 7 (excluding leading-zero cases): $({}^4C_1\times9\times9\times9)-({}^3C_1\times9\times9)=2673$.

Numbers ending in 2 (remainder 2, unit digit 2): $({}^3C_1\times9\times9)-({}^2C_1\times9)=225$. Numbers ending in 7 (remainder 2): $8\times9\times9=648$.

$$P = \frac{225+648}{2673}=\frac{873}{2673}=\frac{97}{297}$$

Solutions · 1