Using the binomial distribution, it is found that the probability that at least 3 customers out of the first 6 will buy bread is of 88.3%.
The formula is:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
In this problem:
The probability that at least 3 customers out of the first 6 will buy bread is given by:
[tex]P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)[/tex]
In which:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 3) = C_{6,3}.(0.65)^{3}.(0.35)^{3} = 0.2355[/tex]
[tex]P(X = 4) = C_{6,4}.(0.65)^{4}.(0.35)^{2} = 0.328[/tex]
[tex]P(X = 5) = C_{6,5}.(0.65)^{5}.(0.35)^{1} = 0.2437[/tex]
[tex]P(X = 6) = C_{6,6}.(0.65)^{6}.(0.35)^{0} = 0.0754[/tex]
Then:
[tex]P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.2355 + 0.328 + 0.2437 + 0.0754 = 0.8826[/tex]
Rounding, the probability is of 88.3%.
More can be learned about the binomial distribution at https://brainly.com/question/24863377