# Probabilities

Printable View

• Jan 16th 2009, 10:30 PM
jax760
Probabilities
Okay I have an answer for this one, but it just seems to easy , Can somone tell me if I did this right. I feel like theres more to solving the problem then what I did. Heres the problem:

The table below gives the probabilities of combinations of religion and political parties in a major U.S city.

Political parites Protestant (A) Catholic (B) Jewish (C) Other(D)

Democrat (E) .35 .10 .03 .02

Republican ( F) .27 .09 .02 .01

Independent (G) .05 .03 .03 .01

What is the probability that a randomly selected person would be a democrat who was not jewish?

The probability I got was .47, all I did was p(.35)+p(.10)+p(.02)= .47
but it just seems to easy.
• Jan 17th 2009, 01:17 AM
mr fantastic
Quote:

Originally Posted by jax760
Okay I have an answer for this one, but it just seems to easy , Can somone tell me if I did this right. I feel like theres more to solving the problem then what I did. Heres the problem:

The table below gives the probabilities of combinations of religion and political parties in a major U.S city.

Political parites Protestant (A) Catholic (B) Jewish (C) Other(D)

Democrat (E) .35 .10 .03 .02

Republican ( F) .27 .09 .02 .01

Independent (G) .05 .03 .03 .01

What is the probability that a randomly selected person would be a democrat who was not jewish?

The probability I got was .47, all I did was p(.35)+p(.10)+p(.02)= .47
but it just seems to easy.

You need the concept of conditional probability: $\displaystyle \Pr(E \, | \, C') = \frac{\Pr(E \cap C')}{\Pr(C')} = \frac{\Pr(E \cap C')}{1 - \Pr(C)}$.

Now substitute the data (you have already unwittingly calculated the numerator: 0.47).

Edit: NO you don't. Your answer is correct. Conditional probability is not relevant in this question.
• Jan 17th 2009, 05:59 AM
awkward
Quote:

Originally Posted by mr fantastic
You need the concept of conditional probability: $\displaystyle \Pr(E \, | \, C') = \frac{\Pr(E \cap C')}{\Pr(C')} = \frac{\Pr(E \cap C')}{1 - \Pr(C)}$.

Now substitute the data (you have already unwittingly calculated the numerator: 0.47).

I guess it's matter of interpretation, but it seems to me the question asks for $\displaystyle \Pr(E \cap C')$, in which case the answer given in the OP is correct.
• Jan 17th 2009, 09:47 AM
Plato
Quote:

Originally Posted by awkward
I guess it's matter of interpretation, but it seems to me the question asks for $\displaystyle \Pr(E \cap C')$, in which case the answer given in the OP is correct.

I agree completely with that.
What would your answer be if the question were “What is the probability that a randomly selected person is a Jewish Democrat?”
• Jan 17th 2009, 09:50 AM
awkward
Quote:

Originally Posted by Plato
I agree completely with that.
What would your answer be if the question were “What is the probability that a randomly selected person is a Jewish Democrat?”

$\displaystyle \Pr(E\cap C) = 0.03$.
• Jan 17th 2009, 10:07 AM
Plato
Quote:

Originally Posted by awkward
$\displaystyle \Pr(E\cap C) = 0.03$.

Again I agree with you. I agreed with you the first time that the OP is correct.
I don't see anything Bayesian about this.
• Jan 17th 2009, 01:48 PM
mr fantastic
Quote:

Originally Posted by Plato
Again I agree with you. I agreed with you the first time that the OP is correct.
I don't see anything Bayesian about this.

Both you guys are right. My mistake.