For the probability question here's my explanation:
For the first ride, total number of options are 3c1 = 3 & number of favorable options is 3c1 = 3 (because he can sit in any car, no preference)
i.e. 3/3 =1
For the second ride, total number of options again are 3c1 = 3 but the number of favorable options is 2c1 = 2 (because he wants to sit in any of the remaining two cars)
i.e. 2/3
For the third ride, total number of options again are 3c1 = 3 & number of favorable options = 1c1 = 1 (because he wants to sit in the only one car left that he has not sat till now)
i.e. 1/3
Hence net probability = 1*2/3*1/3 = 2/9
rishi235 wrote:Hi...
Someone please explain the problems attached herewith.
Thanks...