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Probability: For these problems, consider a standard American deck of playing cards—52 cards, 4 suits in the deck, 13 cards in each suit. Imagine also that the deck has been well shuffled.

1. Draw 1 card. What is the probability of drawing a diamond?

2. Draw 2 cards without replacement. What is the probability that both are diamonds?

3. Draw 2 cards without replacement. What is the probability that the first is a diamond but the second is a club?

4. Draw 2 cards without replacement. What is the probability that the first is a club but the second is a diamond?

5. Draw 2 cards without replacement. What is the probability that one of them is a diamond and the other is a club?

Respuesta :

AL2006
For all probability exercises, the probability is always

         (the number of possible 'successful' results)
divided by  
         (the total possible number of results).

1). Number of diamonds in the deck = 13
     Total cards in the deck  =  52.
     Probability = (13/52) = 1/4 = 25%

2). Probability of the first one being a diamond = 1/4 .
Now there are 51 cards left, and 12 of them are diamonds.
     Probability of the second one being a diamond = 12/51
    
     Probability of both events =
 
                             (1/4) x (12/51) =  3/51 = about 5.88%  (rounded)

3). First one is a diamond:  13/52
Now there are 51 cards left, and 13 of them are clubs.
     Second one is a club:   13/51.

     Probability of both events  =   (13/52) x (13/51)

                                                 =         169/2652  =  about 6.37%  (rounded)

4).  Same logic, same solution, as #3.

5).  6.37% chance that it happens with the diamond coming up first,
6.37% chance that it happens with the club coming up first.
Probability of either one happening  =  12.74% (rounded).

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