On a game show

Question

Has to do with permutations and combinations

 

if you're on a game show that has 15 doors, the contestant gets to choose two doors. Behind two of these doors are expensive cars behind the other 13 there are consolation prizes. The contestant gets to keep both prizes from the doors they pick, what is the probability they get at least one car?

Answer 1

It really doesn't matter the order that they pick the doors as they win both prizes anyway.  Therefore this is a combinations problem.

 

It is actually easier to calculate the compliment probability, the probability that they don't win any cars.  If we subtract that probability from 1, it will give the probability that they choose at least 1 car.

 

There are 15 doors total and you are picking 2 of them.  Therefore, ₁₅C₂ or 105 possible ways to open two of the doors regardless of the result.  This is the total possible outcomes.

 

There are 13 doors with consolation prizes.  Since we don't want cars, we have to pick 2 of them.  Therefore, ₁₃C₂ or 78 possible ways to open the doors and not get a car.

 

P(0 cars) = 78/105 = 26/35

 

Therefore, P(at least 1 car) = 1 - P(0 cars) = 1 - 26/35 = 9/35 or 25.7%

Answer 2

The complement of this event is if the contestant gets two consolation prizes.

 

P(2 Consolation Prizes) = C(13,2)/C(15,2) = 78/105 = 26/35

 

P(At least 1 Car) = 1 - 26/35 = 9/35