rohanberi wrote:A certain roller coaster has 3 cars, and a passenger is equally likely to ride in any 1 of the
3 cars each time that passenger rides the roller coaster. If a certain passenger is to ride
the roller coaster 3 times, what is the probability that the passenger will ride in each of
the 3 cars?

a) 0 b) 1/9 c) 2/9 d) 1/3 e) 1

P(all 3 cars) = P(any car on 1st trip) x P(different car on 2nd trip) x P(different car on 3rd trip)

P(any car on 1st trip) = 1
P(different car on 2nd trip) = 2/3 [at this point, the passenger has occupied one of the cars, which means that 2 of the 3 cars remain not ridden]
P(different car on 3rd trip) = 1/3 [at this point, the passenger has occupied two of the cars, which means that 1 of the 3 cars remains not ridden]

So, P(all 3 cars) = (1) x (2/3) x (1/3) = 2/9

Cheers,
Brent