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

Question

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

A. 0
B. 1/9
C. 2/9
D. 1/3
E. 1

Bunuel · Accepted Answer

First time a passenger can ride any car: p=1;
Second time the passenger should ride another car: p=2/3;
Third time the passenger should ride not the 2 cars he had already driven: p=1/3.

So  P=1*\frac{2}{3}*\frac{1}{3}=\frac{2}{9} .

OR:
Total # of ways the passenger can drive 3 cars (for 3 rides) is 3^3 (for each ride passenger has 3 options);
# of ways the passenger can drive 3 different cars is 3! (ABC, ACB, BAC, BCA, CAB, CBA);

So  P=\frac{3!}{3^3}=\frac{2}{9} .

Answer: C.

Maxirosario2012 · Answer

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?

I couldn´t solve it using the reverse combination approach. Anyone knows how?. But here are the other 3 ways:

Probability approach:
  
First ride: he can choose between 3 cars: 1/3
Second ride: he can choose only (3 less 1) cars: 2/3
Third ride: he can choose the only one in which he hadn´t ride.

Then:
 \frac{1}{3}*\frac{2}{3} *1=\frac{2}{9}

Reverse probability approach:

P = 1-q. 
q = probability that he rides only in one or only in two cars, but not in the three.

q = 1 * \frac{1}{3} * \frac{1}{3}* 3 + 1 * \frac{1}{3} * \frac{1}{3} * 4 = \frac{1}{9} *3 + \frac{1}{9} * 4 = \frac{7}{9}

First term I multiply by 3 because he could use A, B or C. Second term multuply by four because he could use A&B, A&C, C&B or B&C.

P=1-\frac{7}{9}= \frac{2}{9}

Combinatory approach:
  
 C^3_1  *  C^2_1  *  C^1_1  =  3*2*1= 6

Total combinations:  (C^3_1)^3 = 27

\frac{6}{27}=\frac{2}{9}

Ravshonbek · Answer

prob that any car 3/3
prob that any of left 2 cars 2/3
prob of last one 1/3

(3/3)*(2/3)*(1/3)=2/9

Whatever · Answer

My ans is C:

1st ride - prob is 1/3 
2nd ride - prob is 2/3
3rd ride - prob is 1 (as 1 car left ...)

P = 1/2 * 2/3 * 1 = 2/9

x2suresh · Answer

GOOD QUESTION.

SAY 123 are car numbers..
possible combinations 123,132,231,213,321,312

p= 3! * (1/3^3) = 6/27 = 2/9

C

srivas · Answer

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

Soln:
If he is to ride 3 times and since he can choose any of the 3 cars each time, total number of ways is 
= 3 * 3 * 3
= 27

Now the number of ways if he is to choose a different car each time is 
= 3 * 2 * 1
= 6

So the probability is
= 6/27
= 2/9

Ans is  C

lawschoolsearcher · Answer

Everytime the passenger rides a car, he has 3 choices each time. 3 x 3 x 3

To ride 3 times without repeating a car = 3 x 2 x 1

So (3 x 2 x 1)/(3 x 3 x 3) = 2/9

metallicafan · Answer

+1 C

Let's suppose that the cars are A, B, and C.
Now, let's find the probability of picking the cars in this order: A-B-C

\frac{1}{3} * \frac{1}{3} * \frac{1}{3} = \frac{1}{27}

However, there could be more way to organize A-B-C. The numbers of ways is 3! = 6

Then,  \frac{1}{27} * 6 = \frac{6}{27} = \frac{2}{9}

Answer: C

cyberjadugar · Answer

Hi,

Probability = (favorable cases)/(total number of cases)

Total number of ways in which a person can ride car = 3*3*3 = 27
(In first ride he has 3 options to sit, in second right again he has 3 seats available to sit and so on)

Number of favorable cases, i.e., when he rides on different cars;
He can choose seat car in 3 ways in his 1st ride.
He can choose seat car in 2 ways in his 2nd ride.
He can choose seat car in 1 ways in his 3rd ride.
So, 3*2*1 = 6 ways

Thus, probability of choosing different seats = 6/27 = 2/9
(C)

Regards,

ParmarKP · Answer

I did -

For first ride the passenger has to choose one from three, so the probability would be - 1/3
For the next ride he has to choose one out of the two left, so the probability would be - 1/2
And last because he is left with only one ride to choose from to have ridden each once, probability would be - 1/1

Total probability = 1/3*1/2*1 = 1/6

I understood the solution but need to understand why this is incorrect. Thanks.

chetan2u · Answer

HI,
why you are going wrong is because you are not applying the info of Q correctly..

so each time he plans to take the ride he can take any of the three again...
BUT in your solution, you assume that a ride once taken cannot be taken again..

so the solution on your lines would be:-
For first ride the passenger has to choose one from three, so the probability would be - 1/3
For the next ride too, he has to choose one out of the three, so the probability would be - 1/3
And last because he is left with only one ride to choose, but he has to choose again from all three, probability would be - 1/3

Total probability = 1/3*1/3*1/3 = 1/27..
 but within these three the order can differ and there will be 3! ways..
so answer=3!/27=2/9.. 
hope it helps

mvictor · Answer

pretty decent one...
a good question to remind everyone to read everything thoroughly.
we need a passenger to ride all 3 cars.
first try, he can choose any of the cars, so probability is 1.
second try, he can choose any of the two he did not ride, so probability is 2/3
third try, there is only one left out of three, so probability is 1/3
the overall probability:
1 * 2/3 * 1/3 = 2/9

C

if to think logically, we can eliminate right away A and E
among B, C, and D
B 1/9 is the probability if it takes either only 1 car or 1 car first try, then another car two times. so not good
D is way too much, so as well can be eliminated.

BrentGMATPrepNow · Answer

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

sarthak1701 · Answer

Took me a while but let's approach it as a permutation problem. Let's call the cars A, B and C. There are 3P3 = 6 ways to arrange the cars (or 6 ways to ride each of the 3 cars).

Now, the probability of riding each of the cars is 1/3, doing that thrice we get 1/27. Multiply this by the possible ways of riding all the cars and we get 1/27*6 = 2/9.

agrasan · Answer

Instead of going to formula directly, let's think logically on this problem.
Imagine, 3 seats in front of you and the only constraint is  seat has to be different each time .

First pick = 3 seats available; you can choose any so probability = 3/3
Second pick = now you have 2 options out of 3 seats so probability = 2/3
Third pick = now you only have 1 option remaining so probability = 1/3

P = P(First)*P(Second)*P(Third) = (3/3)*(2/3)*(1/3) = 2/9

We multiply because choices are dependent on each other. Again think like you have only have 2 options when you have already taken 1st seat before.

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

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

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