From a pack of 52 cards, we draw one by one, without replacement. If $f(n)$ is the probability that an Ace will appear at the $n^{\text{th}}$ turn, then

Question and options

From a pack of 52 cards, we draw one by one, without replacement. If

f(n)

is the probability that an Ace will appear at the

n^{\text{th}}

turn, then

Correct answer: D - $f(2) = f(3) = \frac{1}{13}$

By symmetry each position is equally likely to hold any card, so P(Ace at nth draw) = 4/52 = 1/13 for every n; f(2) = f(3) = 1/13.