Question: An actuary is studying the prevalence of three health risk factors, denoted by A, B, and C, within a population of women. For each of the three factors, the probability is 0.1 that a woman in the population only has this risk factor (and no others). For any two of three factors, the probability is 0.12 that she has exactly two of these risk factors (but not the other). The probability that a woman has all three risk factors given that she has A and B, is (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?

My attempt: I wrote the "probability that a woman has none of the three risk factors, given that she does not have risk factor A" as Pr(A'andB'andC'|A') as Pr(A'andB'andC'andA')/Pr(A') which just simplifies to Pr(A'andB'andC')/Pr(A') where Pr(A') = (1-.1) = .9. I'm not entirely sure where to go on from there. I also tried to draw a Venn Diagram with three intersecting circles where Pr(AandB'andC') = .1 (same for B and C), but that didn't really get me anywhere The answer is 0.467 (rounded). Can you guys please show me what I'm doing wrong or what I should be doing?

Thank you guys so much!

  • $\begingroup$ Why is $P(A')=1-0.1=.9$? Is $P(A)=0.1$? Note that the probability that a woman only has A is equal to $0.1$. But a woman having $A$ and $B$ still has $A$. $\endgroup$ – megas Dec 20 '14 at 4:59

You can divide the sample into 8 pools. $$ \begin{array}{|c|c|} \hline \text{Factor}& \text{Probability} \\ \hline \text{A} & .1 \\ \hline \text{B} & .1 \\ \hline \text{C} & .1 \\ \hline \text{AB} & .12 \\ \hline \text{AC} & .12 \\ \hline \text{BC} & .12 \\ \hline \text{ABC} & .06 \\ \hline \text{NONE} & .28 \\ \hline \end{array}$$

So you want $$\frac{\text{NONE}}{\text{B}+\text{C}+\text{BC}+\text{NONE}}=\frac{.28}{.1+.1+.12+.28}\approx.467$$

Note, $\text{ABC}=.06$ because for all the pool that have AB, $\frac{2}{3}$ must be AB and $\frac{1}{3}$ must be ABC.


First, lets write down what we know:

  • $ P(A \cap B' \cap C') = P(A' \cap B \cap C')= P(A' \cap B' \cap C) = 0.1$
  • $P(A \cap B \cap C') = P(A \cap B' \cap C)= P(A' \cap B \cap C) = 0.12$
  • $P(A \cap B \cap C|A,B) = \frac{1}{3}.$

As you noted, we are interested in computing the probability $$ P(A' \cap B' \cap C'|A') = \frac{P(A' \cap B' \cap C')}{P(A')}. $$ We have $$ P(A') = 1 - P(A). $$ By the law of total probability, \begin{align} P(A) &= P(A \cap (B \cap C)) + P(A \cap (B' \cap C)) + P(A \cap (B \cap C')) + P(A \cap (B' \cap C')) \\ & = P(A \cap B \cap C) + 0.12 + 0.12 +0.1 \\ & = P(A \cap B \cap C) + 0.34. \end{align} Lets try to determine $P(A \cap B \cap C)$. We have \begin{align} P(A \cap B \cap C) & = P( A \cap B\cap C | A \cap B) \cdot P(A \cap B) = \frac{1}{3} \cdot P(A \cap B). \end{align} Further, by the law of total probability, \begin{align} P(A \cap B) = P(A \cap B \cap C) + P(A \cap B \cap C') = P(A \cap B \cap C) + 0.12. \end{align} Combining the two previous equations, we find \begin{align} P(A \cap B) = \frac{1}{3}P(A \cap B) + 0.12 \quad \Rightarrow \quad P(A \cap B) = \frac{3\cdot 0.12}{2} = 0.18, \end{align} and \begin{align} P(A \cap B \cap C) & = \frac{1}{3} \cdot P(A \cap B) = \frac{1}{3}0.18 = 0.6. \end{align} Returning to the calculation of $P(A)$, we get \begin{align} P(A) = P(A \cap B \cap C) + 0.34 = 0.6 +0.34 = 0.4, \end{align} and in turn \begin{align} P(A') = 1 - P(A) = 1- 0.4 = 0.6. \end{align} Finally, the probability that a woman has no factor can be found by subtracting from $1$ the sum of the probabilities of all disjoint events in which a woman has a factor (exactly one, exactly two, or all three): \begin{align} P(A' \cap B' \cap C') &=1 - \left[P(A \cap B \cap C) + 3\cdot 0.1 + 3 \cdot 0.12 \right]\\ &=.34 - 0.6 = 0.28. \end{align} We now have everything we need to compute the desired result: $$ P(A' \cap B' \cap C'|A') = \frac{P(A' \cap B' \cap C')}{P(A')} =\frac{0.28}{0.6} = 0.4666... $$


there is a mistake cause 0.6+0.34=0.94 thus P(A’)=0.06. and that changes the whole solution. also in the end 1-0.6-0.3-0.36= -0,26?

  • $\begingroup$ Welcome to stackexchange. That is just a typo in the answer from @megas . He or she uses the correct value in what follows, and reaches the correct conclusion. You should delete this (non)answer and edit that one instead. SInce uo not yet have enough reputation to edit freely your changes will be put on a queue to be checked. $\endgroup$ – Ethan Bolker Nov 12 '18 at 16:04

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