Step-by-step explanation:
Total number of ways to arrange the cars
= 6! / (2! * 3!) = 60.
We represent the cars as G, Y and R.
We first arrange the G and Y cars in a row. There are 2 possible cases to investigate.
Case 1: YGY arrangement
The yellow cars are already not stacked side by side.
We have _Y_G_Y_, where the 4 empty spaces are possible places for the red cars to settle. Number of ways = 4C3 = 4.
Case 2: YYG or GYY arrangement
In order for the yellow cars to not be stacked side by side, a red car must be between them.
We have _YRY_G_ or _G_YRY_, where the 3 empty spaces are possible places for the remaining 2 red cars to settle.
Number of ways = 2 * 3C2 = 6.
Hence the probability of such an event
happening is (4 + 6)/60 = 1/6.
The answer is 1/6.
