Probability that sum of 10 numbers randomly picked from a set of size 100 is at least 150

probability combinatorics probability-distributions

A set has 100 numbers:

5x 150
10x 40
5x 30
80x 0

What is the probability that if you take 10 random numbers from this set, with replacement, the sum of the numbers is at least 150?

The only method I can think of is listing out all the combinations of either summing over 150 or summing below 150. Is there a non-brute force way of solving this?


Solution 1:

You can compute this easily using probability generating functions (pgf) and a computer algebra system. The pgf for a single ticket is $$ 0.8x^0+0.05x^{30}+0.1x^{40}+0.05x^{150} $$ Notice how the probability values of the ticket are encoded in the exponents, and the probabilities are the coefficients. You are interested in the sum of several independent tickets. Since the pgf sum of independent random variables is the product of their pgfs, the pgf for the sum of ten tickets is $$ (0.8x^0+0.05x^{30}+0.1x^{40}+0.05x^{150})^{10} $$ If you were to expand out that huge polynomial, you would see the full probability distribution for the sum of ten tickets. Furthermore, if you multiply the above by $1/(1-x)$, and compute the resulting Maclaurin series, the coefficients will encode the cdf of the distribution. Therefore, we just need to compute $$ 1-\left(\text{the coefficient of $x^{149}$ in }\frac{(0.8x^0+0.05x^{30}+0.1x^{40}+0.05x^{150})^{10}}{1-x}\right) $$ Using Mathematica or Wolfram|Alpha with the command

1 - SeriesCoefficient[
((80/100)x^0 + (5/100)x^30 + (10/100)x^40 + (5/100)x^150)^10/(1-x), 
 {x,0,149}],

we find the probability of winning is exactly $0.4251411456$ Wolfram|Alpha computaiton.

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