SOLUTION: At a local fundraising, you can purchase a ticket for $5, and select 3 numbers in any order from 1 to 20. The fundraisers also select 3 numbers. If you have all 3 winning numbers,

Question 1026225: At a local fundraising, you can purchase a ticket for $5, and select 3 numbers in any order from 1 to 20. The fundraisers also select 3 numbers.
If you have all 3 winning numbers, then you win $50.
If you have exactly 2 winning numbers, then you win $10.
And finally, if you have exactly 1 winning number, then you win $5.
What is the probability of net gain in reduced fraction for 3 winning numbers, and 2 winning numbers?
Can someone help me set up this problem? I was able to find the amount of net gain but not the probability of net gain.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
assuming that you can pick the same number 3 times, then the probability of picking the right number for each guess is the same for all 3 guesses.

for example:

333 is a winning number.

under that scenario, then you have a binomial probability type situation.

the formula for that is:

p(x) = c(n,x) * p^x * q^(n-x).

x is the number of times you guessed the right number.
p is the probability of guessing the right number each time.
q is the probability of not guessing the right number each time.
c(n,x) is the number of ways you can guess right x times out of n guesses.

the probability of guessing a number from 1 to 20 is equal to 1/20 = .05

the probability of not guessing a number from 1 to 20 is equal to 19/20 = .95.

the c(n,x) formula means n! / (x! * (n-x)!)

n is equal to 3 because there are 3 guesses.

x is equal to 0,1,2,3 because you can guess right 0 times, 1 time, 2 times, or 3 times.

c(n,x) is the number of times you can get 0 right out of 3, 1 right out of 3, 2 right out of 3, or 3 right out of 3.

using this formula, you get:

c(3,0) = 1
c(3,1) = 3
c(3,2) = 3
c(3,3) = 1

the binomial formula gets you:

p(0) = c(3,0) * p^0 * q^3 = 1 * .05^0 * .95^3 = 0.857375
p(1) = c(3,1) * p^1 * q^2 = 3 * .05^1 * .95^2 = 0.135375
p(2) = c(3,2) * p^2 * q^1 = 3 * .05^2 * .95^1 = 0.007125
p(3) = c(3,3) * p^3 * q^0 = 1 * .05^3 * .95^0 = 0.000125

add up all the probabilities and they equal 1, as they should.

the probability of getting all 3 numbers right is .000125.

the probability of getting exactly 2 numbers right is .007125

you buy the ticket so you lost 5 dollars, right off the bat.

if you get 3 right, then you win 50 dollars, but that only happens .000125 of the time.

your expected value of getting all 3 right is therefore -5 + .000125 * 50 = -4.99375 dollars.

if you get 2 right, then you win 10 dollars, but that only happens .007125 of the time.

your expected value of getting exactly 2 right is therefore -5 + .007125 * 10 = -4.92875.

what does this mean.

if you played the game 1 million times, you could expect to pay 1 million * 5 = 5 million dollars.

you would win 50 dollars .000125 * 1 million = 125 times.

you would therefore win 125 * 50 = 6250 dollars.

you paid 5 million dollars and you won 6250 dollars.

you would have lost 4,993,750 dollars.

divide that by 1 million and it comes out to losing an average of 4.99375 on each ticket.

that's what the expected value calculation gave you.

-4.99375 expected value means you lost 4.99375 dollars on each ticket, assuming that you played the game an infinite number of times.

i chose 1 million, because that gave me an integer result.

your total expected value would be:

-5 + .135375 * 1 + .007125 * 10 + .000125 * 50 = -4.787125

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