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Indylucas
A game show has 4 doors, of which the contestant must choose two. Behind two are cars, behind the other two are consolation. Find the probability the contestant wins exactly one car, no cars, and at least one car. I really need to know how the number of ways P(exactly one car picked)
I'd say when choosing the first door the probability to get a car is 50% (P=0.5). Then we have 3 doors left, so the probability to pick one with no car is 2/3 (P=2/3).. multiply these two and you have the probability of both events happening, which is 1/3.
There are 2 ways of winning exactly one car: (1) A car is picked on the first choice and no car is picked on the second choice: P(1 car) = 2/4 * 2/3 = 4/12 = 1/3 (2) No car is picked on the first choice and a car is picked on the second choice: P(1 car) = 2/4 * 2/3 = 4/12 = 1/3 Events (1) and (2) are mutually exclusive, therefore the probability of (1) or (2) occurring is the sum of their probabilities: 1/3 + 1/3 = 2/3