An actuary is studying the prevalence of three health risk factors, denoted by $A, B, \text{and}\ C$, within a population of women. For each of the three factors, the probability is $0.1$ that a woman in the population has only this risk factor (and no others). For any two of the three factors, the probability is $0.12$ that she has exactly these two risk factors (but not the other). The probability that a woman has all three risk factors, given that she has A and B, is $\frac{1}{3}$
What is the probability that a woman has none of the three risk factors, given that she does not have risk factor $A$?
I started by translating the problem statement into math. We have:
- $P(A \cap B^c \cap C^c) = P(A^c \cap B \cap C^c) = P(A^c \cap B^c \cap C) = 0.1$
- $P(A \cap B \cap C^c) = P(A \cap B^c \cap C) = P(A^c \cap B \cap C) = 0.12$
- $P(A \cap B \cap C \vert A \cap B) = \frac{1}{3}$
We want: $$P(A^c \cap B^c \cap C^c \vert A^c)$$
I've also surmised that: $$P(A \cup B \cup C) = P(A \cap B \cap C^c) + P(A \cap B^c \cap C) + P(A^c \cap B \cap C) - P(A \cap B^c \cap C^c) - P(A^c \cap B \cap C^c) - P(A^c \cap B^c \cap C) + P(A \cap B \cap C) = 0.36 - 0.3 + P(A \cap B \cap C) = 0.06 + P(A \cap B \cap C)$$
Where $$P(A^c \cap B^c \cap C^c) = 1 - P(A \cup B \cup C)$$
Now, from here I wanted to find a more usable form for equation $3$ above in order to potentially determine the value of $P(A \cap B \cap C)$ so that I can determine $P(A \cup B \cup C)$ and thus $P(A^c \cap B^c \cap C^c)$ and finally $P(A^c \cap B^c \cap C^c \vert A^c)$. To that end I've noted the relevant definitions
- $P(A \vert B) = \frac{P(A \cap B)}{P(B)}$
- $P(A_1 \cap A_2 \cap \cdots \cap A_n) = P(A_n \vert A_1 \cap A_2 \cap \cdots \cap A_{n-1})$
The problem I am running into in expressing equation $3$ in a more explicit form is that I don't seem to be getting anywhere with it. For instance, this is my work so far $$P(A \cap B \cap C \vert A \cap B) = \frac{P((A \cap B \cap C) \cap (A \cap B))}{P(A \cap B)} = \frac{P(A \cap B \cap C \cap A \cap B)}{P(A \cap B)} = \frac{P(A \cap B \cap C)}{P(A \cap B)} = \frac{P(A \cap B \cap C)}{P(A) + P(B) - P(A \cup B)}$$
So I have $2$ main questions
- Is my translation of problem statement into math correct?
- If so, is my approach (seeking to find the value for $P(A \cap B \cap C)$ in order to find the other values mentioned correct? And can you offer any hints/guidance as to how I should proceed? I am sure there is a simpler and more elegant way to go about solving this problem but this is the way that my mind has approach it I suppose.
You may find $\mathbb P(A\cap B\cap C)$ using the definition of conditional probability, and the Law of Total Probability.
$$\begin{align}\rm\mathbb P(A\cap B\cap C\mid A\cap B) &=\rm\dfrac{~\mathbb P(A\cap B\cap C)~}{\mathbb P(A\cap B)}\\[1ex] &=\rm \dfrac{\mathbb P(A\cap B\cap C)}{~\mathbb P(A\cap B\cap C)+\mathbb P(A\cap B\cap C^{\small\complement})~}\\[2ex]\therefore\qquad\rm\Bbb P(A\cap B\cap C) &=\rm\dfrac{\mathbb P(A\cap B\cap C\mid A\cap B)~\Bbb P(A\cap B\cap C^{\small\complement})}{1-\mathbb P(A\cap B\cap C\mid A\cap B)}\end{align}$$