AAPL wrote: ↑Fri Jul 22, 2022 4:22 pm

**GMAT Prep**
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

OA

C

**APPROACH #1: Probability rules**
P(rides in all 3 cars) = P(1st car is ANY car

** AND ** 2nd car is different from 1st car

**AND** 3rd car is different from 1st and 2nd cars)

= P(1st car is ANY car)

** x** P(2nd car is different from 1st car)

**x** P(3rd car is different from 1st and 2nd cars)

= 3/3

**x** 2/3

**x** 1/3

= 2/9

Answer: C

**APPROACH #2: Counting techniques**
P(3 different cars) =

**(# of ways to ride in 3 different cars)**/

**(total # of ways to take 3 rides)**
**total # of ways to take 3 rides**
For the 1st ride, there are 3 options

For the 2nd ride, there are 3 options

For the 3rd ride, there are 3 options

So, the total number of ways to take three rides = (3)(3)(3) =

**27**
**# of ways to ride in 3 different cars**
Let the cars be Car A, Car B and Car C

In how many different ways can we order cars A, B and C (e.g., ABC, CAB, BAC, etc)?

**Rule: We can arrange n unique objects in n! ways. **
So, we can arrange 3 unique cars in 3! ways ( =

**6** ways)

P(3 different cars) =

**(# of ways to ride in 3 different cars)**/

**(total # of ways to take 3 rides)**
=

**6**/

**27**
= 2/9

= C

Cheers,

Brent