Question

if you draw two cards from a standard deck of 52 cards without replacement, find the probabilities given below. express answers as fractions.
a. p(king first, jack second) type your answer...
b. p(face card first, ace second) type your answer...
c. p(2 aces) type your answer...

Explanation:

Part a: P(King first, Jack second)

Step 1: Probability of King first

A standard deck has 4 Kings. Total cards = 52. So, \( P(\text{King first}) = \frac{4}{52} \).

Step 2: Probability of Jack second (without replacement)

After drawing a King, 51 cards remain. There are 4 Jacks. So, \( P(\text{Jack second} | \text{King first}) = \frac{4}{51} \).

Step 3: Multiply the probabilities

\( P(\text{King first, Jack second}) = \frac{4}{52} \times \frac{4}{51} = \frac{16}{2652} = \frac{4}{663} \).

Part b: P(face card first, ace second)

Step 1: Probability of face card first

Face cards: Jack, Queen, King. 3 per suit, 4 suits. So, 12 face cards. \( P(\text{face card first}) = \frac{12}{52} \).

Step 2: Probability of ace second (without replacement)

After drawing a face card, 51 cards remain. There are 4 aces. So, \( P(\text{ace second} | \text{face card first}) = \frac{4}{51} \).

Step 3: Multiply the probabilities

\( P(\text{face card first, ace second}) = \frac{12}{52} \times \frac{4}{51} = \frac{48}{2652} = \frac{4}{221} \).

Part c: P(2 aces)

Step 1: Probability of first ace

There are 4 aces. Total cards = 52. So, \( P(\text{first ace}) = \frac{4}{52} \).

Step 2: Probability of second ace (without replacement)

After drawing one ace, 3 aces remain and 51 cards remain. So, \( P(\text{second ace} | \text{first ace}) = \frac{3}{51} \).

Step 3: Multiply the probabilities

\( P(\text{2 aces}) = \frac{4}{52} \times \frac{3}{51} = \frac{12}{2652} = \frac{1}{221} \).

Answer:

s:
a. \(\frac{4}{663}\)
b. \(\frac{4}{221}\)
c. \(\frac{1}{221}\)