Question

In a​ region, there is a 0.9 probability chance that a randomly selected person of the population has brown eyes. Assume 11 people are randomly selected. Complete parts​ (a) through​ (d) below.

a. Find the probability that all of the selected people have brown eyes.

The probability that all of the 11 selected people have brown eyes is:

​(Round to three decimal places as​ needed.)

b. Find the probability that exactly 10 of the selected people have brown eyes.

The probability that exactly 10 of the selected people have brown eyes is:

c. Find the probability that the number of selected people that have brown eyes is 9 or more:

d. If 11 people are randomly​ selected, is it unusual for 9 or more to have brown​ eyes?

Answer 1

Solution:

Here, we have to use binomial distribution with n = 11 and p = 0.9.

Binomial formula is given as below:

P(X=x) = nCx*p^x*q^(n – x)

Where, q =1 – p

We have

n = 11

p = 0.9

q = 1 – 0.9 = 0.1

Part a

Here, we have to find P(X=11)

P(X=x) = nCx*p^x*q^(n – x)

P(X=11) = 11C11*0.9^11*0.1^0

P(X=11) = 1*0.9^11*1

P(X=11)= 0.313811

Required probability = 0.313811

Part b

Here, we have to find P(X=10)

P(X=x) = nCx*p^x*q^(n – x)

P(X=10) = 11C10*0.9^10*0.1^1

P(X=10) = 11*0.9^10*0.1

P(X=10)= 0.383546

Required probability = 0.383546

Part c

Here, we have to find P(X≥9)

P(X≥9) = P(X=9) + P(X=10) + P(X=11)

P(X=x) = nCx*p^x*q^(n – x)

P(X=9) = 11C9*0.9^9*0.1^2

P(X=9) = 55*0.9^9*0.1^2

P(X=9) = 0.213081

From above parts, we have

P(X=10)= 0.383546

P(X=11)= 0.313811

P(X≥9) = P(X=9) + P(X=10) + P(X=11)

P(X≥9) = 0.213081 + 0.383546 + 0.313811

P(X≥9) = 0.910438

Part d

If 11 people are randomly selected, it is not unusual for 9 or more to have brown eyes, because corresponding probability P(X≥9) = 0.910438 is greater than 0.05.

We say it is unusual if the corresponding probability is less than 0.05.