Thread: Greatly appreciate help with this

1. Greatly appreciate help with this

Hi this question has troubled me recently and I would greatly appreciate any help or solutions. I have a fair idea but am generally unsure. For part one I used bayes theorem and the prevalence rate .08 as a figure ( Is this right?) And for part two I am unsure what to use or how to do it. I would greatly appreciate if you could help me out with it. It goes..

For women aged 40-65, the prevalence rate of breast cancer is 800 pet 100,000. A particular
diagnostic test is accurate for 85% of women with breast cancer and for 95% of women
who do not have breast cancer.
(i) The diagnostic test is positive for a particular woman. What is the probability that
she has breast cancer?
(ii) Given a sample of n women, let X denote the number of women who have breats
cancer. State what the probability distribution of this random variable is, and justify
(iii) If 10,000 women were tested, how many would you expect to have breast cancer?

2. Re: Greatly appreciate help with this

Originally Posted by MathJack
For women aged 40-65, the prevalence rate of breast cancer is 800 pet 100,000. A particular
diagnostic test is accurate for 85% of women with breast cancer and for 95% of women
who do not have breast cancer.
What do we know? Write it down!
Define $A$ as the event of a positive breast cancer test.
Define $B$ as the event of having breast cancer.

$P(B) = \tfrac{800}{100000} = 0.008$

$P(A|B) = 0.85$

$P(A^C|B^C) = 0.95$

Originally Posted by MathJack
(i) The diagnostic test is positive for a particular woman. What is the probability that
she has breast cancer?
What does part (i) ask you to find?
It asks us to find $P(B|A)$. Use Bayes.

$P(B|A) = \dfrac{P(A|B) P(B)}{P(A|B) P(B) + P(A|B^C) P(B^C)}$

$P(A|B^C) = 1-P(A^C|B^C) = 1-0.95=0.05$
$P(B^C) = 1-P(B)=1-0.008=0.992$

You have the information to solve the problem.