Respuesta :
There are 17 movies in all. 3 movies can be chosen in 17C3 = 680 ways
2 action movies may be chosen out of 3 action movies in 3C2 = 3 ways
1 comedy movie out of 5 comedy movies may be chosen in 5C1 = 5 ways
P( 2 action movies and 1 comedy movie) = (3C2)(5C1) /17C3 = (3)(5) /680 = 3/136
2 action movies may be chosen out of 3 action movies in 3C2 = 3 ways
1 comedy movie out of 5 comedy movies may be chosen in 5C1 = 5 ways
P( 2 action movies and 1 comedy movie) = (3C2)(5C1) /17C3 = (3)(5) /680 = 3/136
The solution would be like this for this specific problem (please see attached file):
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