Given a percentage what is the probability an event will occur?

Question

The workers' union at a particular university is quite strong. About 94% of all workers employed by the university belong to the workers' union. Recently, the workers went on strike, and now a local TV station plans to interview 4 workers (chosen at random) at the university to get their opinions on the strike. What is the probability that exactly 2 of the workers interviewed are union members?

Answer 1

Use binomial distribution to solve this problem. You can read it in detail here.

http://mathworld.wolfram.com/BinomialDistribution.html

Essentially, you calculate the binomial coefficient of the number of tests, n, against the number of desired successes, k. I'll use

BC(n,k)

as a convenient way to designate a binomial coefficient. You can read more about binomial coefficients here.

http://mathworld.wolfram.com/BinomialCoefficient.html

You take the result and multiply it by the probability of successes, P, taken to nth power. Then you multiply that by the probability of the failures, p, taken to the (n-k)th power.

BC(n,k) * P^k * p^(n-k)

n = 4
--- The number of tests is specified as 4 random workers.

k = 2
--- The desired number of workers we want as union members.

P = 94% or 94/100
--- The probability of a worker being union.

p = 1-P = 6% or 6/100
--- The probability of a worker being non-union.

So plugging all the numbers in, we get

BC(4,2) * (94/100)^2 * (6/100)^(4-2)
= 6 * 0.8836 * 0.0036 ≈ 0.019

So the percentage probability of exactly picking two union workers out of a random selection of four people would be 1.9%.

If you're still confused about how to use binomial distribution, check out my answer to another question which also makes use of it.

http://answers.yahoo.com/question/index;_ylt=AqQhxUxA2wcgfiJ76ZNtseDsy6IX?qid=20061204211610AAsWsWB&show=7#profile-info-1790781967f1e55f8072876884fa12f2aa

Answer 2

The answer depends on the total number of workers. A percentage doesn't tell you that. From the total number of union and non-union workers you could then figure out probabilities based on the changes in permutations of the odds as each worker was selected at random. For example, for ease of calculations, let's say there are 100 workers total - 94 are union and 6 are non-union. The first selection then means there are 94:6 odds that they will select one union worker. Once that is over, however, the odds change! Now there is a 93:6 odds that they will select another union worker. But our percentages do not tell exact numbers! So suppose since the numbers are divisible by 2, there *could* be only 50 total workers! That would mean that the odds change more dramatically as each worker is selected.

This is why I say there is no solution to this problem except that the initial odds of truly random selection of 50-50 union and non-union representation is extremely unlikely.

Answer 3

Uhhh...0.3? Duh? 1.0-0.7=0.3 Common sense: So rare, it's a SUPER POWER!!!