A researcher wishes to conduct a study of the color preferences of new car buyers. Suppose that 40% of this population prefers the color brown. If 18 buyers are randomly selected, what is the probability that exactly a third of the buyers would prefer brown? Round your answer to four decimal places.
Accepted Solution
A:
Answer:0.1655Step-by-step explanation:This is a binomial probability.THe formula is:[tex]nCr*p^r*(1-p)^{n-r}[/tex]Where n is the total number of peopler is how many we want (we want 1/3rd of 18 which is 6, r= 6)p is probability of success (0.4)Putting the information, we get:[tex]nCr*p^r*(1-p)^{n-r}\\=18C6*(0.4)^{6}(0.6)^{12}\\=0.1655[/tex]Thus, the probability that exactly a third of the buyers would prefer brown is 0.1655