Question

you are dealt a hand of three cards, one at a time. find the probability of each of the following.
a) the first club you get is the third card dealt.
b) your cards are all hearts.
c) you get no aces.
d) you have at least one spade.

Explanation:

Step1: Calculate total number of cards

A standard deck has \(n = 52\) cards.

Step2: Part (a)

Number of non - club cards: \(52-13 = 39\)
Probability that first card is non - club: \(P_1=\frac{39}{52}\)
Probability that second card is non - club (after one non - club is drawn): \(P_2=\frac{38}{51}\)
Probability that third card is club (after two non - clubs are drawn): \(P_3=\frac{13}{50}\)
Using the multiplication rule for independent events (in the sense of sequential drawing without replacement), \(P=\frac{39}{52}\times\frac{38}{51}\times\frac{13}{50}=\frac{39\times38\times13}{52\times51\times50}=\frac{741}{5100}=\frac{247}{1700}\approx0.145\)

Step3: Part (b)

Number of hearts: \(13\)
Probability first card is heart: \(P_1=\frac{13}{52}\)
Probability second card is heart (after one heart is drawn): \(P_2=\frac{12}{51}\)
Probability third card is heart (after two hearts are drawn): \(P_3=\frac{11}{50}\)
\(P=\frac{13}{52}\times\frac{12}{51}\times\frac{11}{50}=\frac{13\times12\times11}{52\times51\times50}=\frac{1716}{132600}=\frac{11}{850}\approx0.013\)

Step4: Part (c)

Number of non - aces: \(52 - 4=48\)
Probability first card is non - ace: \(P_1=\frac{48}{52}\)
Probability second card is non - ace (after one non - ace is drawn): \(P_2=\frac{47}{51}\)
Probability third card is non - ace (after two non - aces are drawn): \(P_3=\frac{46}{50}\)
\(P=\frac{48}{52}\times\frac{47}{51}\times\frac{46}{50}=\frac{48\times47\times46}{52\times51\times50}=\frac{103776}{132600}=\frac{8648}{11050}\approx0.783\)

Step5: Part (d)

Probability of no spades:
Number of non - spades: \(52 - 13 = 39\)
Probability first card is non - spade: \(P_1=\frac{39}{52}\)
Probability second card is non - spade (after one non - spade is drawn): \(P_2=\frac{38}{51}\)
Probability third card is non - spade (after two non - spades are drawn): \(P_3=\frac{37}{50}\)
\(P(\text{no spades})=\frac{39}{52}\times\frac{38}{51}\times\frac{37}{50}=\frac{39\times38\times37}{52\times51\times50}=\frac{54834}{132600}=\frac{9139}{22100}\approx0.413\)
Using the complement rule, \(P(\text{at least one spade})=1 - P(\text{no spades})=1-\frac{9139}{22100}=\frac{22100 - 9139}{22100}=\frac{12961}{22100}\approx0.586\)

Answer:

a) \(\frac{247}{1700}\approx0.145\)
b) \(\frac{11}{850}\approx0.013\)
c) \(\frac{8648}{11050}\approx0.783\)
d) \(\frac{12961}{22100}\approx0.586\)