Question #205111

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)

d) P (2 Aces)

b) P (Red card and 4 of spades)

e) P (Queen of hearts and King)

c) P (Spade and Ace of hearts)

f) P (2 of the same card)

20) Find the same probabilities for problem #19 but this time, the card is not placed back in the deck before the 2nd card is drawn. 


Expert's answer

1.

a)

P=14⋅14=116P=\frac{1}{4}\cdot\frac{1}{4}=\frac{1}{16}


d)

P=113⋅113=1169P=\frac{1}{13}\cdot\frac{1}{13}=\frac{1}{169}


b)

P=12⋅152=1104P=\frac{1}{2}\cdot\frac{1}{52}=\frac{1}{104}


e)

P=152⋅113=1676P=\frac{1}{52}\cdot\frac{1}{13}=\frac{1}{676}


c)

P=14⋅152=1208P=\frac{1}{4}\cdot\frac{1}{52}=\frac{1}{208}


f)

P=152⋅152=12704P=\frac{1}{52}\cdot\frac{1}{52}=\frac{1}{2704}


2.

a)

P=14⋅1351=13204P=\frac{1}{4}\cdot\frac{13}{51}=\frac{13}{204}


d)

P=113⋅351=1221P=\frac{1}{13}\cdot\frac{3}{51}=\frac{1}{221}


b)

P=12⋅151=1102P=\frac{1}{2}\cdot\frac{1}{51}=\frac{1}{102}


e)

P=152⋅451=1663P=\frac{1}{52}\cdot\frac{4}{51}=\frac{1}{663}


c)

P=14⋅151=1204P=\frac{1}{4}\cdot\frac{1}{51}=\frac{1}{204}


f)

P=0P=0


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