Question

A television game has 6 shows doors, of which the contests must pick 2. behind two of the doors are expensive cars, and behind the other 4 doors are consolation prizes. Find the probability that the contestant wins exactly 1 car? no car? or atleast one car?

Answer 1

For there to be 1 car, we consider two possible outcomes:
The first door opened has a car or the second door opened has a car.
P(1 car) = 2/6 x 4/5 + 4/6 x 2/5
P(1 car) = 8/15

For there to be no car in either door
P(no car) = 4/6 x 3/5
P(no car) = 2/5

Probability of at least one car is the sum of the probability of one car and probability of two cars:
P(2 cars) = 2/6 x 1/5 
= 1/15
P(1 car) + P(2 cars) = 8/15 + 1/15
= 3/5