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: \((C^3_1)^3 = 27\)

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

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