# See below. Not enough spaceA group of cars consists of 9 Fords and 6 Fiats. A lottery contest plans to give alway 3 randomly to a winner. What is the chance that: 2 of them are fords? ii) And what...

See below. Not enough space

A group of cars consists of 9 Fords and 6 Fiats. A lottery contest plans to give alway 3 randomly to a winner.

What is the chance that: 2 of them are fords?

ii) And what is the expected # of Fiats'

Thanks!

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### 1 Answer

There are `((15),(3))=455` ways to give away 3 cars.

(that is, "15 choose 3" or `(15!)/((3!)(12!))`)

If we have exactly 2 fords, then we have exactly 1 fiat

There are `((9),(2))=36` different pairs of fords we could pick, and `((6),(1))=6` different fiats we could pick.

That means there are:

`((9),(2)) ((6),(1)) = 216` different combinations of cars that would give the winner exactly 2 fords and 1 fiat

So there is a `(216)/(455)` chance of the winner receiving exactly two fords

Now, there are 15 cars with 6 fiats. So if the winner received only 1 car, we would expect the number of fiats received to be `(6)/(15)`

The winner is receiving 3 cars, so we would expect 3 times as many fiats, so:

`((6)/(15))*3=1.2`

PS: in general, to find the expected value, you take the number, multiplied by the probability of that number, and add all of these products together. So for us, you take the number of fiats won multiplied by the probability that you win that number of fiats. Do this for each number and take the sum. So in our situation:

(0 fiats)*(probability of 0 fiats) + (1 fiat)*(probability of 1 fiat) + (2 fiats)*(probability of 2 fiats) + (3 fiats)*(probability of 3 fiats)

This is, (using what we did from before):

`(0)( ((9),(3)) ((6),(0)) )/( 455 ) + (1) ((216)/(455)) + (2)( ((9),(1)) ((6),(2)) )/( 455 ) + (3)( ((9),(0)) ((6),(3)) )/( 455 )`

`=0 + (216)/(455) + (2) (135)/(455)+ (3) (20)/(455) = (546)/(455)=1.2`