a cetain roller coaster has 3 carts, and a passenger is

Question

a cetain roller coaster has 3 carts, and a passenger is equally likely to ride in any 1 of the 3 carts each time the passenger rides. 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 carts?

0

1/9

2/9

1/3

1

lanter1 · Answer

C.

3/3 * 2/3 * 1/3 = 6/27 or 2/9

vidyasagar · Answer

3 carts: A, B, C

NO. OF POSSIBILITIES: A-B-C OR B-C-A OR C-A-B


PROBABILITY: 1/3*1/3*1/3 (+) 1/3*1/3*1/3 (+) 1/3*1/3*1/3

 =1/9

vshaunak@gmail.com · Answer

You have missed three cases - ACB, BAC and CBA.
So probability will be 6/27 = 2/9

KillerSquirrel · Answer

the probability for choosing any cart out of 3 = 3/3 = 1

the probability for choosing an unchosen cart (two left) out of three = 2/3

the probability for choosing the last unchosen cart out of three = 1/3

since all those events are dependent then:

3/3*2/3*1/3 = 6/27 = 2/9

Dek · Answer

C. 2/9

Let the carts be A,B & C

First ride he chooses A = 1/3
Second ride he chooses B = 1/3
Third ride he chooses C = 1/3

So all three happening i.e A, B, and C
= 1/3*1/3*1/3 = 1/27

Now ABC can be arranged in 3! ways
3! = 6

So 1/27 * 6 = 2/9

andrehaui · Answer

I believe that is 2/9 because:

1*(1-1/3)*(1-2/3)

1= 3 cars
1-1/3 =2 cars
1-2/3 =1 car

a cetain roller coaster has 3 carts, and a passenger is

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