SOLUTION: draw one card (out of a deck of 52) at a time until you get four cards. count the number of jacks you get. find the probability distribution

Question 667107: draw one card (out of a deck of 52) at a time until you get four cards. count the number of jacks you get. find the probability distribution
Answer by Edwin McCravy(20054) (Show Source): You can put this solution on YOUR website!

You will either get 0 jacks, 1 jack, 2 jacks, 3 jacks, or 4 jacks.
Calculate the probabilities of each of the 5 possibilities.

P(0 jacks)

There are 48 non-jacks. so we can choose them C(48,4) ways.
The denominator is C(52,4) So the probability is  = 0.718737

P(1 jack)

We can choose the jack any of C(4,1) ways, we can choose the
3 non-jacks any of C(48,3) ways.
The denominator is C(52,4) So the probability is AMP Parsing Error of [("C(4,1)""C(48,3)")/"C(52,4)"]: Invalid expression. Closing bracket expected "C(48,3)")/"C(52,4)" at /home/ichudov/project_locations/algebra.com/templates/Algebra/Expression.pm line 187.
.
 = 0.255551

P(2 jacks)

We can choose the 2 jacks any of C(4,2) ways, we can choose the
2 non-jacks any of C(48,2) ways.
The denominator is C(52,4) So the probability is AMP Parsing Error of [("C(4,2)""C(48,2)")/"C(52,4)"]: Invalid expression. Closing bracket expected "C(48,2)")/"C(52,4)" at /home/ichudov/project_locations/algebra.com/templates/Algebra/Expression.pm line 187.
.
 =  0.025000

P(3 jacks)

We can choose the 3 jacks any of C(4,3) ways, we can choose the
1 non-jack any of C(48,1) ways.
The denominator is C(52,4) So the probability is AMP Parsing Error of [("C(4,3)""C(48,1)")/"C(52,4)"]: Invalid expression. Closing bracket expected "C(48,1)")/"C(52,4)" at /home/ichudov/project_locations/algebra.com/templates/Algebra/Expression.pm line 187.
.
 = 0.000709

P(4 jacks)

We can choose the 4 jacks any of C(4,4) ways.
The denominator is C(52,4) So the probability is  = 0.000004

Probability distribution:

X = the number of jacks:

 X      P(X)
 0   0.718737
 1   0.255551
 2   0.025000
 3   0.000710
 4   0.000004
-------------
     1.000002

The total of the probabilities would be 1 if 
the probabilities weren't rounded off. 

Edwin

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