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saviop
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


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
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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
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saviop
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

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
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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
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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
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+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
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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,
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nakib77
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

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.

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.
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ParmarKP

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.


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

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
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nakib77
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

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.
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nakib77
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
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
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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.
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