Atelevision game has 6 shows doors, of which the contests must pick 2. behind two of the doors are expensive cars, and behind the other 4 doors are consolation prizes. find the probability that the contestant wins exactly 1 car? no car? or atleast one car?

In this question, there are 4 consolation door and 2 car doors of total 6 doors. That means for the first door the chance to get consolation is 4/6 and car is 2/6. In second door, only 5 remains so the chance to get consolation is either 4/5 (if 1st door car) or 3/5(if 1st door consolation)

Win 1 car condition: 
1. 1st door car, 2nd door consolation = 2/6 * 4/5 =8/30
2. 1st door consolation, 2nd door car= 4/6 * 2/5 = 8/30
8/30+8/30= 16/30

No car condition
1. 1st door consolation, 2nd door consolation= 4/6 * 3/5= 12/30

At least one car condition
1. 1st door car, 2nd door car = 2/6 * 1/5 =2/30
2. 1st door car, 2nd door consolation = 2/6 * 4/5 =8/30
3. 1st door consolation, 2nd door car= 4/6 * 2/5 = 8/30
2/30 + 8/30 + 8/30= 18/30

4+2=-2. 1+2=3. y=-2/3x+b. 2=4/3+b. 2/3=b. the answer is y= -2/3x+2/3

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