From a single deck of cards you select a card look at it put

From a single deck of cards, you select a card, look at it, put it back in the deck, and shuffle. If you do this 10 times, what is the probability of drawing exactly (1) one ace, (b) two aces, (c) three aces, (d) four aces (each of these is calculated separately)? Out of the 10 draws of cards, what is the mean number of aces that you would expect to find?

What would be the probability of finding three or four aces out of the 10 draws? How is this calculation different than the ones above?

Try to do a similar problem with face cards.

When might this type of issue come up in real life? Certainly, we don\'t deal with cards often!

Solution

A)

P(one ace) = 4/52 = 1/13

B)

P(two aces) = (4/52)(4/52) = 1/169

C)

P(three aces) = (4/52)(4/52)(4/52) = 1/2197

D)

P(4 aces) = (4/52)(4/52)(4/52)(4/52) = 1/28561

E)

mean = n p

As p = 4/52 = 1/13, n = 10,

mean = 10(1/13) = 10/13 [ANSWER]

F)

What would be the probability of finding three or four aces out of the 10 draws? How is this calculation different than the ones above?

Note that P(between x1 and x2) = P(at most x2) - P(at most x1 - 1)          
          
Here,          
          
x1 =    3      
x2 =    4      
          
Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    10      
p = the probability of a success =    0.076923077      
          
Then          
          
P(at most    2   ) =    0.963773376
P(at most    4   ) =    0.999512006
          
Thus,          
          
P(between 3 and 4) =    0.03573863   [ANSWER]  

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These probabilities may come up if you are a gambler.