Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards. The probability distribution of number of aces is given by:

Drawing 2 cards with replacement from a deck of 52 cards. After picking the first card, it's returned to the deck before the second draw.

Total cards in deck = 52

Number of aces = 4

Number of non-aces = 48

For any single draw:

P(getting an ace) = $\frac{4}{52} = \frac{1}{13}$

P(not getting an ace) = $\frac{48}{52} = \frac{12}{13}$

---

For X = 0 (no aces), both cards must be non-aces.

P(X = 0) = P(1st card is not ace) × P(2nd card is not ace)

$= \frac{12}{13} \times \frac{12}{13}$

$= \frac{144}{169}$

---

For X = 1 (exactly one ace), there are two ways:

First card is ace, second card is non-ace:

$= \frac{1}{13} \times \frac{12}{13} = \frac{12}{169}$

First card is non-ace, second card is ace:

$= \frac{12}{13} \times \frac{1}{13} = \frac{12}{169}$

P(X = 1) = $\frac{12}{169} + \frac{12}{169}$

$= \frac{24}{169}$

---

For X = 2 (both aces), both cards must be aces.

P(X = 2) = P(1st card is ace) × P(2nd card is ace)

$= \frac{1}{13} \times \frac{1}{13}$

$= \frac{1}{169}$

---

The probability distribution is:

x	0	1	2
P(x)	$\frac{144}{169}$	$\frac{24}{169}$	$\frac{1}{169}$

Therefore, the answer is Option 3.