Two cards are drawn at random, one after the other, without replacement from a well-shuffled pack of 52 playing cards. The probability that the first card is a spade and the second card is an ace is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{1}{52}\)
The number of aces left for the second draw depends on whether the first spade is the ace of spades, so the first card is split into two cases that cannot happen together.
The first card is the ace of spades: then 3 aces are left among 51 cards. By the multiplication theorem:
\( \displaystyle P_1 = \frac{1}{52}\times\frac{3}{51} = \frac{3}{2652} \)
The first card is one of the other 12 spades: then all 4 aces are left among 51 cards:
\( \displaystyle P_2 = \frac{12}{52}\times\frac{4}{51} = \frac{48}{2652} \)
Adding the two cases:
\( \displaystyle P = \frac{3+48}{2652} = \frac{51}{2652} = \frac{1}{52} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{1}{51}\) — Incorrect. This is \(\frac{13}{52}\times\frac{4}{51}\), which wrongly keeps all 4 aces even when the first card is the ace of spades.
Option (2) \(\displaystyle\frac{1}{68}\) — Incorrect. This is \(\frac{13}{52}\times\frac{3}{51}\), which wrongly removes an ace every time.
Option (4) \(\displaystyle\frac{1}{884}\) — Incorrect. This is \(\frac{1}{52}\times\frac{3}{51}\), only the case in which the first card is the ace of spades.