Based on data from the national center of health, 26% of adults are overweight. A random sample of 15 adults is taken. Calculate the probability that three of four of 15 people selected are overweight?
This is an example of a binomial event and probability distribution. For each person you select, they are either overweight or not-only two possible outcomes. If we define 'success' as selecting an overweight person, the probability of success when we select someone is denoted p, and is given as 26%, or, in decimal form, .26.
p=.26 <-- probability of success
q=1-.26=.74 <-- probability of 'failure', choosing someone who is not overweight. p+q=1
n=15 <-- Number of people we are selecting
We need to add the probabilities of getting exactly 3 out of 15 and getting exactly 4 out of 15. We'll use x to represent the number of successes in our selection of 15 adults:
Our expression of the total probability of selecting either 3 or 4 people out of 15: P(x=3) + P(x=4)
The binomial probability formula: P(X=x) = nCx * px * qn-x
nCx is our notation for a combination, which mathematically equals [ n! / ( x! * (n-x)! ) ]
The probability of picking 3 or four people out of 15 is the sum of the following two expressions:
P(X=3) = 15C3 * (0.26)3 * (0.74)15-3 = 455 * (0.26)3 * (0.74)12 = 0.216
P(X=4) = 15C4 * (0.26)4 * (0.74)15-4 = 1365 * (0.26)4 * (0.74)11 = 0.227
P(X=3) + P(X=4) = 0.216 + 0.227 = 0.443
The probability of selecting three or four overweight people in a random sample of 15 is 0.443.