ruchisharma wrote:A certain roller coaster had 3 cars, and a passenger is equally to ride in any 1 od 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
Let's rewrite the probability.
P(ride in all 3 cars)
= P(ride in any car on 1st ride
AND ride in an unridden car on the 2nd ride
AND ride in the last unridden car on the 3rd ride)
= P(ride in any car on 1st ride)
x P(ride in an unridden car on the 2nd ride)
x P(ride in the last unridden car on the 3rd ride)
Aside: P(ride in any car on 1st ride) = 1 (pick any car)
P(ride in an unridden car on the 2nd ride)=2/3 (there are 3 cars to choose from, but only 2 of them have not yet been ridden)
P(ride in the last unridden car on the 3rd ride)=1/3 (3 cars to choose from, but we have ridden in 2 of them, leaving only 1 to ride)
So, P(ride in any car on 1st ride)
x P(ride in an unridden car on the 2nd ride)
x P(ride in the last unridden car on the 3rd ride) = 1
x 2/3
x 1/3
= 2/9
