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^(th) turn, then Option 1: f(2) = 1/13 > f(3) Option 2: 1/13 > f(2) > f(3) Option 3: f(3) > f(2) = 1/13 Option 4: f(2) = f(3) = 1/13

Answer: f(2) = f(3) = 1/13. Solution: f(n) is the probability that the n^(th) card drawn is an Ace. Think of it this way: before drawing, every card is equally likely to be in any position. The n^(th) position is no more or less likely to hold an Ace than any other position. By symmetry: f(n) = 4/52 = 1/13 for all n Let's verify with f(2). The 2^(nd) card can be an Ace in two ways: Case 1: 1^(st) card was an Ace AND 2^(nd) card is also an Ace = 4/52 x 3/51 Case 2: 1^(st) card was not an Ace AND 2^(nd) card is an Ace = 48/52 x 4/51 Adding them: f(2) = (4 x 3)/(52 x 51) + (48 x 4)/(52 x 51) = (12 + 192)/2652 = 204/2652 = 1/13 ✓ This confirms that the position of the draw doesn't matter — each position is equally likely to hold an Ace. f(1) = f(2) = f(3) = *s = f(52) = 1/13

Modern Math

Probability

Hard

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

$f(2) = \frac{1}{13} > f(3)$

$\frac{1}{13} > f(2) > f(3)$

$f(3) > f(2) = \frac{1}{13}$

$f(2) = f(3) = \frac{1}{13}$

✅ Correct Option: 4

$f(n)$ is the probability that the $n^{\text{th}}$ card drawn is an Ace.

Think of it this way: before drawing, every card is equally likely to be in any position. The $n^{\text{th}}$ position is no more or less likely to hold an Ace than any other position.

By symmetry:

$f(n) = \dfrac{4}{52} = \dfrac{1}{13}$ for all $n$

Let's verify with $f(2)$ . The $2^{\text{nd}}$ card can be an Ace in two ways:

Case 1: $1^{\text{st}}$ card was an Ace AND $2^{\text{nd}}$ card is also an Ace

$= \dfrac{4}{52} \times \dfrac{3}{51}$

Case 2: $1^{\text{st}}$ card was not an Ace AND $2^{\text{nd}}$ card is an Ace

$= \dfrac{48}{52} \times \dfrac{4}{51}$

Adding them:

$f(2) = \dfrac{4 \times 3}{52 \times 51} + \dfrac{48 \times 4}{52 \times 51}$

$= \dfrac{12 + 192}{2652}$

$= \dfrac{204}{2652}$

$= \dfrac{1}{13}$ ✓

This confirms that the position of the draw doesn't matter — each position is equally likely to hold an Ace.

$f(1) = f(2) = f(3) = \cdots = f(52) = \dfrac{1}{13}$

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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