Question 1178552: There are 131 passengers waiting to board a plane. 74 passengers have two carry on items,48 have one carry on item and 9 have no carry on item. TSA people randomly select 3 passengers to search. Find the probability that:
A. All 3 have on carry on
B. 1 had one carry on, and 2 have two carry ons
C.1 has one carry on, 1 had two carry ons and 1 has no carry on.
Explain
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! number of ways to choose 3 from 131 is 131C3, the denominator.
Numerator is 48C3 for a.
probability is 0.0472
another way (48/131)(47/130)(46/129)=0.0472
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numerator is 48C1*74C2; probability is =0.3541
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numerator ice 74*48*9; probability is 0.0873
74C1*48C1*9C1/131C3
the left side coefficients of the C in the numerator adds to the left side in the denominator, and the right side, too.
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