JEE PYQ: Probability Question 14
Question 14 - 2021 (26 Feb Shift 2)
A seven digit number is formed using digits 3, 3, 4, 4, 4, 5, 5. The probability, that number so formed is divisible by 2, is:
(1) $\frac{6}{7}$
(2) $\frac{4}{7}$
(3) $\frac{3}{7}$
(4) $\frac{1}{7}$
Show Answer
Answer: (3)
Solution
Total arrangements $= \frac{7!}{2!3!2!}$. For divisibility by 2, last digit must be 4. Fix 4 at end: $\frac{6!}{2!2!2!}$. $P = \frac{6!/(2!2!2!)}{7!/(2!3!2!)} = \frac{90}{210} = \frac{3}{7}$.
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