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?	Choices
A) 0  B) 1/9  C) 2/9  D) 1/3  E) 1
Probability = (favorable cases)/(total number of cases)   Total number of ways in which a person can ride car = 3*3*3= 
<gadget id="calculator">3*3*3</gadget>
<output>27</output>
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   Answer: 2/9
<result>C</result>
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?