Question 1048830: PLEASE HELP! I NEED THIS ANSWER PLEASE!!!!!!
In a lottery game, a player picks six numbers from 1 to 48. If 4 of those 6 numbers match those drawn, the player wins third prize. What is the probability of winning this prize? (Give your answer as a fraction.)
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the probability of getting any one number right is 1/48.
assuming that the same number can be drawn more than once, then the same probability exists for all 6 numbers.
this looks like a binomial probabilty where p(x) = p^x * q^(n-x) * c(n,x)
p is equal to 1/48
q is equal to 47/48
x goes from 0 to 1.
p(0) = (1/48)^0 * (47/48)^6 * c(6,0)
p(1) = (1/48)^1 * (47/48)^5 * c(6,1)
p(2) = (1/48)^2 * (47/48)^4 * c(6,2)
p(3) = (1/48)^3 * (47/48)^3 * c(6,3)
p(4) = (1/48)^4 * (47/48)^2 * c(6,4)
p(5) = (1/48)^5 * (47/48)^1 * c(6,5)
p(6) = (1/48)^6 * (47/48)^0 * c(6,6)
player needs to get 4 out of 6.
that would be p(4) = (1/48)^4 * (47/48)^2 * c(6,4)
the result of that would be equal to 2.709190541 * 10^(-6)
that's the same as .000002709190541
the total probability has to be equal to 1.
the following excel spreadsheet display shows that this is true and also confirms the answer given above.
c(n,x) is the combination formula that is equal to n! / (x! * (n-x)!)
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