Answer:
0.000773 is the probability that atleast 12 out of 14 will have brown eyes.
Step-by-step explanation:
We are given the following information:
We treat people having brown eyes as a success.
P(people have brown eyes) = 41% = 0.41
Then the number of people follows a binomial distribution, where
[tex]P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}[/tex]
where n is the total number of observations, x is the number of success, p is the probability of success.
Now, we are given n = 14
We have to evaluate:
[tex]P(x \geq 12) = P(x = 12) + P(x = 13) + P(X = 14) \\= \binom{14}{12}(0.41)^{12}(1-0.41)^2 + \binom{14}{13}(0.41)^{13}(1-0.41)^1 + \binom{14}{14}(0.41)^{14}(1-0.41)^0\\= 0.0007 + 0.00007 + 0.000003\\= 0.000773[/tex]
0.000773 is the probability that atleast 12 out of 14 will have brown eyes.
Yes, it is an unusual event due to small probability values.
