# Probability Question!

#### isabellamercer5681

##### New Member
The probability that a person in the United States has dark brown eye color is 18%.
Three unrelated people in the United States are selected at random.
a)
What is the probability that all three have
dark brown eye colour
?
b)
What is the probability that at least
one person has dark brown eye colour? (Hint: use
a complement rule)

B says complement rule but to simply do 1.0-0.82=0.18 seems too simple to be correct and doesn't account for the fact that there is 3 people so I'm not sure if it should be that or 0.18x0.82x0.82???

#### obh

##### Well-Known Member
Hi Isabella,
The probability that at least one person has a dark brown eye color:
1 - the probability that no one has a dark brown eye color:

So what is the probability that no one has a dark brown eye color?

#### isabellamercer5681

##### New Member
Hi Isabella,
The probability that at least one person has a dark brown eye color:
1 - the probability that no one has a dark brown eye color:

So what is the probability that no one has a dark brown eye color?
I did the complement rule so that 0.82 is the probability of neither having dark brown eyes.

do I then have to do (0.18)(0.82)(0.82)=0.12 ??

#### obh

##### Well-Known Member
Hi Isabella,

Let's do it step by step.

The probability that a person in the United States has dark brown eye color is 0.18

1. What is the probability that a person in the United States doesn't have dark brown eye color?
2. What is the probability that no one of the 3 students has a dark brown eye color?

Last edited:

#### isabellamercer5681

##### New Member
Hi Isabella,

Let's do it step by step.

The probability that a person in the United States has dark brown eye color is 0.18

1. What is the probability that a person in the United States doesn't have dark brown eye color?
2. What is the probability that no one of the 3 students has a dark brown eye color?
1. 0.82 or 82%
2. (0.82)(0.82)(0.82)=0.551

#### obh

##### Well-Known Member
Great.

So the probability no one of the 3 students has a dark brown eye color is 0.551
Now, what is the probability that at least one will have brown eye color?

#### isabellamercer5681

##### New Member
Great.

So the probability no one of the 3 students has a dark brown eye color is 0.551
Now, what is the probability that at least one will have brown eye color?
(0.18)(0.82)(0.82)=0.121

#### obh

##### Well-Known Member
Option1: no one of the 3 students has a dark brown eye color.
Option2: at least one person has a dark brown eye color.

Can you imagine option3?

#### isabellamercer5681

##### New Member
Option1: no one of the 3 students has a dark brown eye color.
Option2: at least one person has a dark brown eye color.

Can you imagine option3?
if all 3 had dark brown eye colour

#### obh

##### Well-Known Member
No, Option2 includes the option that all the 3 had dark brown eye colour

#### Dason

(0.18)(0.82)(0.82)=0.121
That looks like the probability that first person has it and the other two don't. What about when the second person has it? Or the third?

#### obh

##### Well-Known Member
Option1: no one of the 3 students has a dark brown eye color.
Option2: at least one person has a dark brown eye color.

Option1 + option2 contains all the possibilities so P(options1)+p(option2)=1.

Usually, the easier way to understand is to write all the options
for simple writing:
1 - dark brown eye
0 - other eyes color

Following all options:
Option, student-1, student-2, student-3
1, 0,0,0
2, 1,0,0
3, 0,1,0
4, 0,0,1
5, 1,1,0
6, 0,1,1
7, 1,0,1
8, 1,1,1

The question is: What is the probability that at least one person has a dark brown eye color?

So the is the probability that one of the following options will happen: 2,3,4,5,6,7,8

You calculated only option 2. you need to add options 3,4,5,6,7,8

You may do it the hard way p(2)+p(3)+...p(8)
Or you may use the hint, as you know that p(1)+p(2)+...+p(8)=1.
I suggest that you will try to calculate on both ways to improve your intuition

Cheers.

PS, you may look also the binomial distribution, saving calculations using combinatorics,