# Thread: Conditional probability.

1. ## 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

2. Hi

OK for 1. and 2.

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

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

3. hi,

How does the 7% come in to the answer?

is the answer 0.00719?

thanks

4. 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]
Originally Posted by running-gag
Hi

OK for 1. and 2.

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

$p(B) = p(B|A) p(A) + p(B|\overline{A}) p(\overline{A})$
I think Q3 only requires $\Pr(B \cap A) = \Pr(B \, | \, A) \Pr(A) = \, ....$

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