Suppose a certain disease has four causes, A, B, C, and D. They can occur simultaneously and independently. Any one of them is enough to cause the disease. First, if you know that a patient has the disease, what is the probability of it being caused by A or B (or both)? For that, I reasoned as follows:

Assuming F denotes the probability of having the disease:

$$P(A\cup B|F) = \frac {P((A \cup B) \cap F)}{P(F)}$$

By commutativity,

$$ \frac {P((A \cup B) \cap F)}{P(F)} = \frac {P((A\cap F) \cup (B \cap F)}{P(F)}$$

Since any of the factors is enough to cause the disease, $P(A \cap F)$ = P(A) and the same for B. Therefore,

$$\frac {P((A\cap F) \cup (B \cap F)}{P(F)} = \frac{P(A \cup B)}{P(F)} $$

Using the independence property,

$$ \frac{P(A \cup B)}{P(F)} = \frac{P(A) + P(B) - P(A)P(B)}{P(F)} $$

Using the fact that $P(F) = 1 - P(A^c)P(B^c)P(C^c)P(D^c)$ the answer follows.

Now suppose that causes C and D are mutually exclusive. What is the probability of a disease being caused by C, given that you know the patient has the disease? My idea is this:

First, $$P(C|F) = \frac{P(C \cap F)}{P(F)} = \frac{P(C)}{P(F)}$$

The probability of getting the disease in the previous question can also be written as:

$$P(F) = P(A \cap B \cap C \cap D) + P(A^c \cap B \cap C \cap D) + P(A \cap B^c \cap C \cap D) ...$$

and so on, with $2^4 - 1$ combinations (the -1 for the combination where all the factors are negative). Since now C and D are mutually exclusive, all the terms which contain their intersection are no longer in the sample space (have probability 0). I can thus get the new probability of failure by subtracting those terms from the previously calculated probability of failure.

$$P(F_{new}) = P(F_{old}) - P(A^c \cap B \cap C \cap D) - P(A \cap B^c \cap C \cap D) - P(A^c \cap B^c \cap C \cap D) - P(A \cap B \cap C \cap D)$$

The same idea holds for P(C). Since P(C) can be expressed as the sum of the probabilities of all the possible intersections with C (2^3 of them), the probability of C with mutual exclusivity is:

$$P(C_{new}) = P(C_{old}) - P(A \cap B \cap C \cap D) - P(A^c \cap B^c \cap C \cap D) - P(A^c \cap B \cap C \cap D) - P(A \cap B^c \cap C \cap D)$$

The answer would then be:

$$\frac {P(C_{new})}{P(F_{new})}$$

Did I miss anything?



1 Answer 1


Sorry if this is too late an answer...

Part (a) I agree.

Part (b): I think the intention is to treat it as a somewhat separate question to part (a). That the probabilities $P(A), P(B), P(C), P(D)$ here are to be treated as though they take into account the mutual exclusivity of $C$ and $D$. So I don't think we need to calculate $P(C_{new})$ - it is just simply $P(C)$.

So: \begin{eqnarray*} P(C|F) &=& \dfrac{P(C \cap F)}{P(F)} \\ &=& \dfrac{P(C)}{P(F)} \end{eqnarray*}

\begin{eqnarray*} P(F) &=& 1 - P(A^c \cap B^c \cap C^c \cap D^c) \\ &=& 1 - P(A^c)P(B^c)P(C^c \cap D^c) \\ \end{eqnarray*}

\begin{eqnarray*} P(C^c \cap D^c) &=& P((C \cup D)^c) \\ &=& 1 - P(C \cup D) \\ &=& 1 - P(C) - P(D) + P(C \cap D) \\ &=& 1 - P(C) - P(D)\qquad\mbox{since $C$ and $D$ mutually exclusive} \\ \end{eqnarray*}

Therefore, \begin{eqnarray*} P(C|F) &=& \dfrac{P(C)}{1 - (1 - P(A))(1 - P(B))(1 - P(C) - P(D))} \end{eqnarray*}


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