Women carrying a certain gene are ten times more likely to develop breast cancer. Only 1 out of 100 women carries this gene. If a woman has breast cancer, what is the probability that she carries this gene?
Carrying the gene $ P(C) = 1/100 = 0.01$
Probability of woman who has not the gene developing the cancer $P(D) = x$
Probability of woman who has the gene developing the cancer $P(D | C) = 10x$
I suppose that the question asks $P(C | D) = ?$
Using bayesian formula we can find the answer
$$P(C|D)= \frac{P(D|C)*P(C)}{P(D)}$$
$$P(C|D)= \frac{10x*0.01}{x} = 0.1$$
Is this solution right?
Thanks in advance.
The first thing is that the probability of a woman who does not have the gene developing the cancer is actually $P(D|\overline{C})=x$, where $P(\overline{C})=1-P(C)=0.99$
Your application of Bayes' rule looks right in terms of the formula, however you need to proceed as follows:- $$\begin{align}P(C|D)&= \frac{P(D|C)*P(C)}{P(D)}\\&= \frac{P(D|C)*P(C)}{P(D|\overline{C})P(\overline{C})+P(D|C)P(C)}\\&=\frac{10x\times0.01}{(x\times0.99)+(10x\times0.01)}=0.092\end{align}$$