A television game show has 11 doors, of which the contestant must pick 3. Behind 3 of the doors are expensive cars, and behind

Question

2 years ago

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A television game show has 11 doors, of which the contestant must pick 3. Behind 3 of the doors are expensive cars, and behind

the other 8 doors are consolation prizes. The contestant gets to keep the items behind the 3 doors she selects. Determine the probability that the contestant wins at least one car.

Mathematics

1 answer:

bezimeni [28] · Answer 1 · 2022-04-16T12:56:09+03:00

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

A television game show has 11 ​doors, of which the contestant must pick 3. Behind 3 of the doors are expensive​ cars, and behind

A television game show has 11 doors, of which the contestant must pick 3. Behind 3 of the doors are expensive cars, and behind