JEE PYQ: Probability Question 12
Question 12 - 2021 (25 Feb Shift 2)
Let $A$ be a set of all 4-digit natural numbers whose exactly one digit is 7. Then the probability that a randomly chosen element of $A$ leaves remainder 2 when divided by 5 is:
(1) $\frac{1}{5}$
(2) $\frac{2}{9}$
(3) $\frac{97}{297}$
(4) $\frac{122}{297}$
Show Answer
Answer: (3)
Solution
Total cases: $(4 \times 9 \times 9 \times 9) - (3 \times 9 \times 9)$… Computing $n(A)$: 4-digit numbers with exactly one 7. Favourable: those ending in 2 or 7 with remainder 2 mod 5. By careful counting: probability $= \frac{97}{297}$.
BETA
