SOLUTION: The percentage of Americans who are vegetarians is 3.2%. Four people are randomly selected and are asked if they are vegetarians. Find the probability that at most one person said

Algebra ->  Permutations -> SOLUTION: The percentage of Americans who are vegetarians is 3.2%. Four people are randomly selected and are asked if they are vegetarians. Find the probability that at most one person said       Log On


   

Question 913000: The percentage of Americans who are vegetarians is 3.2%. Four people are randomly selected and are asked if they are vegetarians. Find the probability that at most one person said they are vegetarians. Hint: You must find the probability that zero people (x = 0) and one person (x = 1) said they are vegetarians.
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
Using P+%28x%29=+highlight_green%28nCx%29%28p%5Ex%29%28q%29%5E%28n-x%29+
p and q are the probabilities of success and failure respectively.
In this case p = .032 and q = .968, n = 4
nCx+=+%28n%21%29%2Fx%21%28n+-+x%29%21%29
P(x ≤ 1) = P(0) + P(1) = (.968)^4 + 4C1 (.032)^1(.968)^2
Using TI
P(x ≤ 1) = binomcdf(4, .032, 1)