# How can i solve this probability question?

• May 29th 2009, 08:14 PM
change_for_better
How can i solve this probability question?
In a certain town 45% of the people have black hair,30% have brown eyes and 20% have both black hair and brown eyes.if a person is selected at random from the town,

What is the probability that he has brown eyes and doesn't have black hair?
• May 29th 2009, 10:12 PM
matheagle
Let A=black hair, B=brown eyes.

P(A)=.45, P(B)=.3 and P(AB)=.2.

You want $P(A'B)$, where the prime denotes the complement.

I would use $P(A'B)+P(AB)=P(B)$.

SO $P(A'B)+.2=.3$, GIVING YOU $P(A'B)=.1$.

My question is where is the probability that this person is male?
• May 29th 2009, 10:25 PM
change_for_better
Quote:

Originally Posted by matheagle
Let A=black hair, B=brown eyes.

P(A)=.45, P(B)=.3 and P(AB)=.2.

You want $P(A'B)$, where the prime denotes the complement.

I would use $P(A'B)+P(AB)=P(B)$.

My question is where is the probability that this person is male?

probability=0.3/0.45=0.6

Is my answer true?
• May 29th 2009, 10:34 PM
matheagle
Quote:

Originally Posted by change_for_better
probability=0.3/0.45=0.6

Is my answer true?

I don't agree at all, but if you wanted the conditional probabilities that would be ${.2\over .45}$ and ${.2\over .3}$

BUT NOT ${.3\over .45}$.

And this is not a conditional (to me) since you have 'brown eyes and doesn't have black hair.'

AND implies intersection and not a given.
• May 29th 2009, 10:39 PM
change_for_better
Quote:

Originally Posted by matheagle
I don't agree at all, but if you wanted the conditional probabilities that would be ${.2\over .45}$ and ${.2\over .3}$

BUT NOT ${.3\over .45}$.

And this is not a conditional (to me) since you have 'brown eyes and doesn't have black hair.

AND implies intersection and not a given.

should I use :

P(A intersection B)=P(A)*P(B/A)??
• May 29th 2009, 10:41 PM
matheagle
why?
20% have both black hair and brown eyes
means, to me, P(AB)=.2.

I left out the basic subtraction earlier today.
I expected you to finish, but I am tired and I just went back and finished the problem for you.
All I wanted, was for you to subtract .2 from .3.
• May 29th 2009, 10:50 PM
change_for_better
Quote:

Originally Posted by matheagle
why?
20% have both black hair and brown eyes
means, to me, P(AB)=.2.

I left out the basic subtraction earlier today.
I expected you to finish, but I am tired and I just went back and finished the problem for you.
All I wanted, was for you to subtract .2 from .3.

I think the answer will be :

P(A)-P(A intersection B)=0.45-0.2=0.25

Is my answer now true?
• May 29th 2009, 10:54 PM
change_for_better
Quote:

Originally Posted by matheagle
why?
20% have both black hair and brown eyes
means, to me, P(AB)=.2.

I left out the basic subtraction earlier today.
I expected you to finish, but I am tired and I just went back and finished the problem for you.
All I wanted, was for you to subtract .2 from .3.

sorry,I know my mistake now ..

I'm very sorry..
• May 30th 2009, 06:45 PM
matheagle
Quote:

Originally Posted by change_for_better
probability=0.3/0.45=0.6

Is my answer true?

and by the way, what is 30/45?
• May 31st 2009, 12:57 AM
HallsofIvy
Quote:

Originally Posted by change_for_better
In a certain town 45% of the people have black hair,30% have brown eyes and 20% have both black hair and brown eyes.if a person is selected at random from the town,

What is the probability that he has brown eyes and doesn't have black hair?

Here's how I would do that problem. Imagine the town has 1000 residents. Then 30% of 1000= 300 people have brown eyes and 20% of 1000= 200 people have both brown eys and black hair. That leaves 300- 200= 100 who have brown eyes but NOT black hair. The probability of selecting such a person, all people in town being equally likely, is 100/1000= .10.

And, as matheagle pointed out, .3/.45 is NOT .6!
• May 31st 2009, 04:01 AM
mr fantastic
Quote:

Originally Posted by change_for_better
In a certain town 45% of the people have black hair,30% have brown eyes and 20% have both black hair and brown eyes.if a person is selected at random from the town,

What is the probability that he has brown eyes and doesn't have black hair? Mr F says: He .....??

Well, if we're showing different approaches then here's mine.

Make a Karnaugh table (since I'm hopeless with formulas):

$\begin{tabular}{l | c | c | c} & Black hair & Not black hair & \\ \hline Brown eyes & 0.2 & & 0.3 \\ \hline Not brown eyes & & & \\ \hline & 0.45 & & 1 \\ \end{tabular}
$

Now it's like a simple Sudoko puzzle to fill in the rest of the probabilities:

$\begin{tabular}{l | c | c | c} & Black hair & Not black hair & \\ \hline Brown eyes & 0.2 & 0.1 & 0.3 \\ \hline Not brown eyes & 0.35 & & 0.7\\ \hline & 0.45 & 0.55 & 1 \\ \end{tabular}
$

I'm sure you can fill in the last square.

So the answer to the question "What is the probability that [the person] has brown eyes and doesn't have black hair?" is easily read from this table to be 0.1.

However, since no information is given about males and females, the actual question "What is the probability that he has brown eyes and doesn't have black hair?" cannot be answered.