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# A certain roller coaster has 3 cars, and a passenger is

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Manager
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A certain roller coaster has 3 cars, and a passenger is [#permalink]

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21 Sep 2004, 02:58
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

Ans: C. Can anyone explain why? I thought it's B.

If x > 0, then 1/[v(2x)+vx] =

A. 1/v(3x)
B. 1/[2v(2x)]
C. 1/(xv2)
D. (v2-1)/vx
E. (1+v2)/vx

Ans: D. Can't figure out why.

June 25, 1982, fell on a Friday. On which day of the week did June 25, 1987, fall? (Note:1984 was a leap year.)

A. Sunday
B. Monday
C. Tuesday
D. Wednesday
E. Thursday

Ans: C. Can't think of a fast way to solve this.
CEO
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21 Sep 2004, 03:04

Regards
Praetorian
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24 Sep 2004, 06:14
I will answer the probability Q

there are 3 cars and the requirement is that the person rides in 3 different cars

When he rides for the first time it does not matter wihch car he rides So P of riding in any one car for the first time(because there are 3 choices now) is 3/3
Now when he rides second time there are still 3 cars but he should ride in any one of the two cars other than the first one (because there only 2 choices )
so P = 2/3
FInally there are still 3 cars but he can ride only in 1 because he has already used two cars so P = 1/3

total P = 3/3 * 2/3 * 1/3 = 2/9

I hope this helps.

Anand.
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24 Sep 2004, 23:37
Probability Question:

Total number of Pb = 3 x 3 x 3= 9 (each time he has three options and he rides thrice)
Favorable deals = 3 x 2 x 1 (He has three options to use for the first and two for the next, three for the last).
Pb = 6/27 = 2/9
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24 Sep 2004, 23:45
For the calendar problem. Ans is E and not C. http://www.gmatclub.com/phpbb/viewtopic ... ght=friday
24 Sep 2004, 23:45
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