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?
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?
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?
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.
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.
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!
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.