Suppose we uniformly and randomly select a permutation from the 20! Permutations of 1, 2, 3 ,…..,20. What is the probability that 2 appears at an earlier position than any other even number in the selected permutation?
(A) 1/2
(B) 1/10
(C) 9!/20!
(D) Node of the above
Answer: (B)
Explanation: All even numbers have the same probability of being first (The odd numbers do not matter here). And there are 10 of them. The probability 2 comes before the other 9 evens is = 1/10.
So, option (B) is correct.
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