In a game of chance, the probability of winning is 30%. If 15 games are played, find the probability of winning exactly six games.

Answer 1

If x is number of wins, then x follows a binomial distribution

then the probability of x = k, number of wins, for n trials and probability of win p is

n! p^k (1-p)^n-k

----

k! (n-k)!

For our problem, n = 15, k = 6 and p = 0.3

Answer 2

The probability of winning on any given try is 0.30, and that means the probability of not winning is 0.70, because p + q = 1. Let us set the probability of winning as p, and the probability of not winning as q. The binomial expansion of (p + q)ⁿ describes the probability for any number of tries where n is equal to the number of tries, n in this case is equal to 15. Winning exactly six times also means losing exactly 9 times in this case, so we need to determine the p⁶q⁹ term, and we need the coefficient of the binomial expansion to tell us how many ways that can happen.

Completely forget about the binomial expansion for this, only a masochist would attmpt that. Instead use Pascal's triangle to determine binomial coefficients. Expanding Pascal's triangle to the fifteenth power gives 3043 as the coefficient for p⁶q⁹ , so you answer is found by solving for 3043 x (0.3)⁶ x (0.7)⁹ = 0.08952, or 8.95%.

Now, I only did this out of curiosity. This problem is actually just busy work, and the fundamentals of this can be illustrated in a much less complicated fashion.