Question 1191476
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You own 50 DVDs consisting of 25 comedies, 15 dramas, and 10 thrillers. 
You randomly pick 4 movies to watch during a long train ride. 
What is the probability that you pick at least one DVD of each type of movie?
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<pre>
First we calculate the number of all possible combinations of 50 DVDs taken 4 at a time.


It is the number of combinations, of course,  

    {{{C[50]^4}}} = {{{(50*49*48*47)/(1*2*3*4)}}} = 230300.


This number is the cardinality of the space of events in this problem.



Next, we calculate the number of all possible favorable outcomes.


The favorable outcomes are 3 disjoint sets  (2C, 1D, 1T),  (1C, 2D, 1T),  (1C, 1D, 2T),

where "C" means comedy CD;  "D" means drama CD;  "T" means thriller CD.


The number of elements in the set  (2C, 1D, 1T)  is  {{{((25*24)/2)*15*10}}} = 300*15*10 = 45000.

The number of elements in the set  (1C, 2D, 1T)  is  {{{25*((15*14)/2)*10}}} = 25*105*10 = 26250.

The number of elements in the set  (1C, 1D, 2T)  is  {{{25*15*((10*9)/2)}}} = {{{25*15*45}}} = 16875.


So, the total number of the favorable outcomes is  45000 + 26250 + 16875 = 88125.


Finally, the probability under the problem's question is


    P = {{{favorable/total}}} = {{{88125/230300}}} = {{{75/196}}} = 0.3827   (rounded).    <U>ANSWER</U>
</pre>

Solved and explained.



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Hello, &nbsp;this instruction about using the complement formula is &nbsp;IRRELEVANT &nbsp;to the posted problem.