# Conditional probability.

• Mar 3rd 2009, 08:36 AM
bolton_boy
Conditional probability.
Hi,

I have a test tomorrow and I am really struggling with a question:

Suppose that among 100,000 women with negative mammograms, 20 will get breast cancer and that 1 woman in 10 with a positive mammogram will get breast cancer. Let A denote positive mammograms and B breast cancer. Find:

1. p(b|a(compliment))

2. p(b|a)

3. suppose that 7% of the general female population will have a positive mammogram. Find the probability of developing breast cancer among this female population.

I think i understand 1. and 2., would the first part just be 20/100,000 and would part 2 just be 1/10?

however i really don't get the last part?

Thanks
• Mar 3rd 2009, 08:53 AM
running-gag
Hi

OK for 1. and 2.

For 3.
$\displaystyle p(B) = p(B \cap A) + p(B \cap \overline{A})$

$\displaystyle p(B) = p(B|A) p(A) + p(B|\overline{A}) p(\overline{A})$
• Mar 3rd 2009, 09:07 AM
bolton_boy
hi,

How does the 7% come in to the answer?

thanks
• Mar 3rd 2009, 06:42 PM
mr fantastic
Quote:

Originally Posted by bolton_boy
[snip]
Suppose that among 100,000 women with negative mammograms, 20 will get breast cancer and that 1 woman in 10 with a positive mammogram will get breast cancer. Let A denote positive mammograms and B breast cancer. Find:

[snip]

3. suppose that 7% of the general female population will have a positive mammogram. Find the probability of developing breast cancer among this female population.

[snip]

Quote:

Originally Posted by running-gag
Hi

OK for 1. and 2.

For 3.
$\displaystyle p(B) = p(B \cap A) + p(B \cap \overline{A})$

$\displaystyle p(B) = p(B|A) p(A) + p(B|\overline{A}) p(\overline{A})$

I think Q3 only requires $\displaystyle \Pr(B \cap A) = \Pr(B \, | \, A) \Pr(A) = \, ....$
• Mar 4th 2009, 08:54 AM
running-gag
Quote:

Originally Posted by mr fantastic
I think Q3 only requires $\displaystyle \Pr(B \cap A) = \Pr(B \, | \, A) \Pr(A) = \, ....$

Exact ... sorry