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?

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- May 29th 2009, 08:14 PMchange_for_betterHow 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 PMmatheagle
Let A=black hair, B=brown eyes.

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

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

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

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

My question is where is the probability that this person is male? - May 29th 2009, 10:25 PMchange_for_better
- May 29th 2009, 10:34 PMmatheagle
I don't agree at all, but if you wanted the conditional probabilities that would be $\displaystyle {.2\over .45}$ and $\displaystyle {.2\over .3}$

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

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 PMchange_for_better
- May 29th 2009, 10:41 PMmatheagle
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 PMchange_for_better
- May 29th 2009, 10:54 PMchange_for_better
- May 30th 2009, 06:45 PMmatheagle
- May 31st 2009, 12:57 AMHallsofIvy
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 AMmr fantastic
Well, if we're showing different approaches then here's mine.

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

$\displaystyle \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:

$\displaystyle \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.