MATH SOLVE

9 months ago

Q:
# A television 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?

Accepted Solution

A:

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:Β

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

Win 1 car condition:Β

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