Probability?

Question

according to the U.S. census, one-third of all businesses are owned by women. If we select 25 businesses at random, what is the probability that 10 or more of them are owned by women?

Answer 1

Since the census data shows an uneven probability distribution of men vs women business owners, the answer is not simply 10/25. Additionally, the answer will not be 1/3, because the question asks for the probability of finding 10 or more female business owners (at a probability of 1/3, one would expect to obtain 8.333 females from a sample of 25). Thus, the answer will likely be somewhere below 1/3.

Let:

p(w) = probability of choosing a business owned by a woman
p(m) = probability of choosing a business owned by a man

Assuming a business can only be owned by a man or woman (as opposed to being owned by an intangible entity such as a government or parent corporation), and assuming that a man and woman can not co-own a business, then 100% of all businesses are owned by a man OR by a woman:

p(w) + p(m) = 1

We are looking for the probability of 10 or more than 10 of the business owners randomly chosen will be women. The only way I can think of to solve this problem exactly is to use combinatorics. That is, let the order of the chosen business owners be something like:

mwmmwmmwwmwmmmwwmmwwwmwmm

where the first business chosen is owned by a man, the second is owned by a woman, and so on down to the 25th business.

We are only interested in the number of women as opposed to the order in which they are chosen. The only way I can think of to solve this problem exactly is to calculate the probability of getting a number of female business owners on the interval [10,25].

This can be done fairly simply by the binomial expansion theorem, where n=25 (look at the Wikipedia entry listed below). The expansion goes:

p(m)^25 + 25*p(m)^24*p(w) + 300*p(m)^23*p(w)^2 + 2300*p(m)^22*p(w) + ... + 25*p(m)*p(w)^24 + p(w)^25 = 1

There are a total of 26 terms in the expansion. Note that each term denotes the probability of getting a certain number of men or women. The third term, for instance is:

300 * (2/3)^23 * (1/3)^2 = .00297

This means that there is a .297% chance of getting 23 male business owners and only 2 female business owners, given the census data.

The solution, then, is to sum the probabilities of all terms with an exponent for p(w) equal to or greater than 10. Using MS Excel, I calculated this to be:

p(w >= 10) = p(w=10) + p(w=11) + p(w=12) + ... + p(w=25)

or:

p(w >= 10) = SUM(w=10:25) = .3044 or 30.44%

Therefore, there is a 30.44% chance of getting 10 or more women in a sample of 25 business owners.

Answer 2

Since there are many businesses, we can suppose that the probabilty that each selected business is owned by a woman is 1/3 and that such events are pairwise independent. So, we have a binomial distribution, and the probabilty that you have exactly k businesses owned ny women is p(k) = Bin(25, k) (1/3)^k (2/3)^(25 -k). Here Bin(25, k) is the number of simple combinations of 25 choose k.

Since you want the probability that 10 or more businesses are owned by women, the answer is Sum (k=1o, 25) p(k) Ypu can compute the probabiilty using the binomial function in an Excel spreadsheet.