JEE PYQ: Probability Question 2
Question 2 - 2021 (16 Mar Shift 2)
Let A denote the event that a 6-digit integer formed by 0, 1, 2, 3, 4, 5, 6 without repetitions, be divisible by 3. Then probability of event A is equal to:
(1) $\frac{9}{56}$
(2) $\frac{4}{9}$
(3) $\frac{3}{7}$
(4) $\frac{11}{27}$
Show Answer
Answer: (2)
Solution
Total 6-digit numbers: $6 \cdot 6!$. Sum of digits divisible by 3: Case I: ${1,2,3,4,5,6}$: $6!$ ways. Case II: ${0,1,2,4,5,6}$: $5 \cdot 5!$ ways. Case III: ${0,1,2,3,4,5}$: $5 \cdot 5!$ ways. $P = \frac{6! + 2 \cdot 5 \cdot 5!}{6 \cdot 6!} = \frac{720 + 1200}{4320} = \frac{4}{9}$.
BETA
