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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?

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  1. 11 May, 03:47
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    For there to be 1 car, we consider two possible outcomes:

    The first door opened has a car or the second door opened has a car.

    P (1 car) = 2/6 x 4/5 + 4/6 x 2/5

    P (1 car) = 8/15

    For there to be no car in either door

    P (no car) = 4/6 x 3/5

    P (no car) = 2/5

    Probability of at least one car is the sum of the probability of one car and probability of two cars:

    P (2 cars) = 2/6 x 1/5

    = 1/15

    P (1 car) + P (2 cars) = 8/15 + 1/15

    = 3/5
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