Calculating three or four of 15 adults

Question

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?

Answer 1

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) = _nC_x * p^x * q^n-x

_nC_x 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.