This is a lottery odds question. It involves the Mega Millions, where 5 numbers are selected from the intergers 1-56 (without replacement) then a special 6th ball is selected from a new set of intergers 1-46. The odds of winning are 1/175711536. If exactly 200,000,000 random numbers are entered, what is the probability of 2 or fewer winners. I know how to find the prob of 0 winners, but I don't know how to find exactly 1 or exactly 2. Any help would be appreciated.
Let P be the probability of winning = (1/175...). Let n be the number of tickets = 200M. Let x be the exact number of winners.
First, calculate z, the number of ways you could have exactly x winners. It is n-choose-x; the formula is:
z=n!/(n-x)!x!
(where ! means factorial n! 1*2*3...*n. Also, zero!=1)
So if x=0 then z=1. if x=1 then z=n, if x=2 then z=n*(n-1)/2
Probability of exactly x winners is:
z * P^x * (1-P)^(n-x)
I will let a = 1/175,711,536= 5.591145*10^-9 , to save typing.
The expected number, among 200,000,000, is L = 2*10^8 a = 1.13823
The probability of P(k) of k winners is given by Poisson statistics:
P(k) = L^k*e^-L/k!
Probability of 0 winners = 0.32039
Probability of 1 winner = 0.36467
probability of 2 winners = 0.20754
Probability of 3 winners = 0.07874
Probability of 4 winners = 0.02241
Probability of 2 or fewer winners = 89.26%
The Poisson formula I used is different from Economyst's formula, which may be the more exact. The numbers I calculate with Poisson differ by less that 1% in a relative sense. The Poisson formula is easier to use. Try them both and see.
Thanks drwls, and excellent suggestion to use the Poisson. Just for grins, I checked your results with my methodology; they are the same in all cases (at least up to the 5th significant digit)
To calculate the probability of exactly 1 or exactly 2 winners when 200,000,000 random numbers are entered, we need to use the concept of binomial distribution.
In this case, the probability of winning for any single entry is 1/175,711,536. Let's denote this probability as p.
To find the probability of exactly k winners (where k can be 0, 1, or 2), we can use the binomial probability formula:
P(k) = (n choose k) * p^k * (1-p)^(n-k)
Where:
- n is the total number of entries (200,000,000 in this case)
- (n choose k) represents the binomial coefficient, which is calculated as n! / (k! * (n-k)!)
- p^k represents the probability of k winners
- (1-p)^(n-k) represents the probability of (n-k) non-winners
To find the probability of 0 winners, you mentioned you already know how to do it. Let's denote it as P(0).
For exactly 1 winner:
P(1) = (200,000,000 choose 1) * (1/175,711,536)^1 * (1-1/175,711,536)^(200,000,000-1)
For exactly 2 winners:
P(2) = (200,000,000 choose 2) * (1/175,711,536)^2 * (1-1/175,711,536)^(200,000,000-2)
To calculate these probabilities, you can use a calculator, software, or programming language with support for large numbers and combinations.
Finally, to find the probability of 2 or fewer winners, you need to sum up the probabilities of 0, 1, and 2 winners:
P(0 or 1 or 2) = P(0) + P(1) + P(2)
By calculating these probabilities using the binomial distribution formula, you will obtain the desired probability.