Answer:
Step-by-step explanation:
Given that a cheese can be classified as either raw dash milk or pasteurized. Suppose that 91% of cheeses are classified as pasteurized.
Each cheese is independent of the other and there are only two outcomes.
Hence X = no of cheeses pasteurized is binomial (2,0.91)
a) P(x=2 ) when n=2 is[tex](0.91)^2=0.8281[/tex]
b) n=6.
P(x=6) = [tex](0.91)^6=0.5679[/tex]
c) P(atleast one raw dash milk) = 1-P(all pasteurised)
=0.8281
Since probability is 0.83 it is not unusual.