Question 250530:  Suppose that 5 cards are selected with replacement from a standard deck of 52 playing cards. The  
deck contains 13 denominations (ace, two, three, … , ten, jack, queen, king) of each of four suits  
(hearts, diamonds, spades, and clubs). The kings, queens, and jacks are known as face cards. A card  
is selected and its suit and denomination recorded. The card is replaced in the deck. The deck is  
shuffled, and the process repeated 4 more times.  
a. What is the probability that all five cards are face cards? 
 
 
b. What is the probability that at least one card is a face card? 
 
c. What is the probability that exactly one card is a face card? 
 
 
d. What is the probability that no hearts are selected? 
 
e. What is the probability one or fewer hearts are selected?  
 
 Answer by stanbon(75887)      (Show Source): 
You can  put this solution on YOUR website! Suppose that 5 cards are selected with replacement from a standard deck of 52 playing cards. The deck contains 13 denominations (ace, two, three, … , ten, jack, queen, king) of each of four suits (hearts, diamonds, spades, and clubs). The kings, queens, and jacks are known as face cards.  
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A card is selected and its suit and denomination recorded. The card is replaced in the deck. The deck is shuffled, and the process repeated 4 more times. 
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a. What is the probability that all five cards are face cards? 
Since the cards are replaced the selections are independent. 
P(a face card in one draw) = 12/52 
P(5 face cards in 5 draws) = (12/52)^5 
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b. What is the probability that at least one card is a face card? 
P(at least one) = 1 - P(none) 
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P(no face card in 5 draws) = (40/52)^5 
P(at least one face card in 5 draws) = 1 - (40/52)^5 
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c. What is the probability that exactly one card is a face card? 
It's binomial with n= 5 ; p = 12/52, x = 1 
binompdf(5,12/52,1) = 0.4040 
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d. What is the probability that no hearts are selected? 
(3/4)^5 
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e. What is the probability one or fewer hearts are selected? 
binomial with n=5, p= 3/4, 0<= x <=1 
binomcdf(5,3/4,1) = 0.0156 
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Cheers, 
Stan H. 
  
 
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