Question

Two percent of women age $45$ who participate in routine screening have breast cancer. Ninety percent of those with breast cancer have positive mammographies. Eight percent of the women who do not have breast cancer will also have positive mammographies. Given that a woman has positive mammography, what is the probability she has breast cancer?

Short answer

The probability that the woman has breast cancer is$0.1867$

Step-by-step solution

Step 1:Given Information

Given that two percent of women age $45$ who participate in routine screening have breast cancer. Ninety percent of those with breast cancer have positive mammographies. Eight percent of the women who do not have breast cancer will also have positive mammographies.

Explanation

$B$=Breast cancer

$C$=Positive mammographies

$P\left(B\right)=2\mathrm{\%}=0.02$

$P(C\mid B)=90\mathrm{\%}=0.9$

$P\left(C\mid B^c\right)=8\mathrm{\%}=0.08$

Use the complement rule:

$P\left(A^c\right)=P(\mathrm{not}A)=1-P\left(A\right)$

$P\left(B^c\right)=1-P\left(B\right)=1-0.02=0.98$

Explanation of Bayes's Theorem

Use the equation

$P\left(A_i\mid B\right)=\frac{P\left(B\mid A_i\right)P\left(A_i\right)}{\sum _{j=1}^k P\left(B\mid A_j\right)P\left(A_j\right)}$

$P(B\mid C)=\frac{P(C\mid B)P\left(B\right)}{P(C\mid B)P\left(B\right)+P\left(C\mid B^c\right)P\left(B^c\right)}$

Substitute the value,

$=\frac{0.9\times 0.02}{0.9\times 0.02+0.08\times 0.98}$

$=\frac{0.018}{0.018+0.0784}$

$=\frac{0.018}{0.0964}$
$=\frac{180}{964}$

We get,

$=\frac{45}{241}$

localid="1648546616788" $=0.1867$

Final Answer

The probability she has breast cancer is$\frac{45}{241}=0.1867$.