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Probability Question 2

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by ruchisharma » Fri Aug 13, 2010 7:21 am
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
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by ketandoshi » Fri Aug 13, 2010 8:22 am
Is the probability 1/3 ? What I did is :

probability for a passenger to sit in 1 car is = 1/3 and there are 3 cars so probability for a passenger to sit in 3 cars is 1. Then he takes ride 3 times so probability = 1/3 i.e. ( probability for each time / # of times) ........ this is my guess....please correct me if I am wrong.

Thanks
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by Brent@GMATPrepNow » Fri Aug 13, 2010 8:34 am
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
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by selfmade » Fri Aug 13, 2010 8:45 am
This is just great explaination Brent. I liked the way, you simplified the problem first.
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by sanju09 » Sat Aug 14, 2010 1:07 am
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

Each time, there are 3 possible cars for a passenger to ride on; hence, for three trials, there are as many as 3 × 3 × 3 = 27 ways in total.

If, the rider is supposed to ride a different car in each of his three trials, then his first trial can be made in 3 ways, second in 2 ways, and the third in just 1 way, favor totaling to 3 × 2 × 1 = 6 ways, and hence the required probability is

= 6/27 = [spoiler]2/9

C[/spoiler]
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