Question

1. if 2 cards are selected from a standard deck of cards. the first card is placed back in the deck before the second card is drawn. find the following probabilities:

a) p(heart and club)
b) p(red card and 4 of spades)
c) p(spade and ace of hearts)
d) p(2 aces)
e) p(queen of hearts and king)
f) p(2 of the same card)

Explanation:

Step1: Calculate probability of independent events

Since the first card is replaced, the two draws are independent events. The probability of two independent events \(A\) and \(B\) is \(P(A\cap B)=P(A)\times P(B)\). A standard deck has \(n = 52\) cards.

Part a)

• Number of hearts \(n_{H}=13\), number of clubs \(n_{C}=13\)
• \(P(\text{Heart})=\frac{13}{52}\), \(P(\text{Club})=\frac{13}{52}\)
• \(P(\text{Heart and Club})=\frac{13}{52}\times\frac{13}{52}=\frac{169}{2704}=\frac{1}{16}= 0.0625\)

Part b)

• Number of red cards \(n_{R}=26\), number of 4 of spades \(n_{4S}=1\)
• \(P(\text{Red card})=\frac{26}{52}\), \(P(4\text{ of spades})=\frac{1}{52}\)
• \(P(\text{Red card and }4\text{ of spades})=\frac{26}{52}\times\frac{1}{52}=\frac{26}{2704}=\frac{1}{104}\approx0.0096\)

Part c)

• Number of spades \(n_{S}=13\), number of Ace of hearts \(n_{AH}=1\)
• \(P(\text{Spade})=\frac{13}{52}\), \(P(\text{Ace of hearts})=\frac{1}{52}\)
• \(P(\text{Spade and Ace of hearts})=\frac{13}{52}\times\frac{1}{52}=\frac{13}{2704}=\frac{1}{208}\approx0.0048\)

Part d)

• Number of Aces \(n_{A}=4\)
• \(P(\text{Ace first draw})=\frac{4}{52}\), \(P(\text{Ace second draw})=\frac{4}{52}\)
• \(P(2\text{ Aces})=\frac{4}{52}\times\frac{4}{52}=\frac{16}{2704}=\frac{1}{169}\approx0.0059\)

Part e)

• Number of Queen of hearts \(n_{QH}=1\), number of Kings \(n_{K}=4\)
• \(P(\text{Queen of hearts})=\frac{1}{52}\), \(P(\text{King})=\frac{4}{52}\)
• \(P(\text{Queen of hearts and King})=\frac{1}{52}\times\frac{4}{52}=\frac{4}{2704}=\frac{1}{676}\approx0.0015\)

Part f)

• For the first card, we can draw any card (\(P = 1\)). For the second card to be the same as the first, given the first card is replaced, the probability that the second - card matches the first is \(\frac{1}{52}\)
• \(P(2\text{ of the same card})=1\times\frac{1}{52}=\frac{1}{52}\approx0.0192\)

Answer:

a) \(0.0625\)
b) \(\approx0.0096\)
c) \(\approx0.0048\)
d) \(\approx0.0059\)
e) \(\approx0.0015\)
f) \(\approx0.0192\)