Question 1178542: 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 ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website!
Hi
131 Passengers: 74(2). 48(1) AND 9(0) CARRYONS
randomly select 3 passengers
Find the probability that:
A. All 3 have one carry on = 48C3/134C3
B. 1 had one carry on, and 2 have two carryons = 74C2)(48C1)/(134C3)
C.1 has one carry on, 1 had two carry ons and 1 has no carry on.
(74C1)(48C1)(9C1)/(134C3)
Wish You the Best in your Studies.
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