Conditional Probability - Example

Question

I was trying to solve the following problem:

Compute $\mathbb P(C|D)$, given $\mathbb P(D|C) = 0.1$, $\mathbb P(C) = 0.2$, $\mathbb P(D|E) = 0.25$, $\mathbb P(E) = 0.5$, $\mathbb P(D|F) = 0.75$ and $\mathbb P(F) = 0.5$, where $E$ and $F$ are mutually exclusive and exhaustive events (either $E$ happens or $F$ happens, and one of the two must happen).

In order to calculate the unconditional probability of $D$, I used the low of total probability.

Since $C$, $E$, $F$ are the only possible events in which $D$ ocurres, we have:

$P(D)=P(D|C)P(C)+P(D|E)P(E)+P(D|F)P(F)$

$P(D)=0.02+0.125+0.375$

$P(D)=0.52$

It follows that:

$P(C|D)=\frac{P(D|C)P(C)}{P(D)}$

$P(C|D)=\frac{0.02}{0.52}$

$P(C|D)≈0.0385$

Am I right? Intuitively, the solution seems 'rare' to me - i.e. too low $P(C|D)$ in comparison with other probabilities, so I am afraid that I made a mistake somewhere.

Any help is appreciated.

Thanks in advance.

Answer 1

You can only use the law of total probability over a set of exhaustive, mutually exclusive events. In this case E and F are exhaustive and mutually exclusive, so you have $$P(D) = P(D|E)P(E) + P(D|F)P(F).$$ The remainder of your argument is correct.

Answer 2

The probability for a union of mutually exclusive events is the sum of the probabilities for those events.

The claim is that $E,F$ are mutually exclusive and exhaustive events.

So $C$ cannot be mutually exclusive with $E\cup F$.   Indeed, because $E,F$ are exhaustive, their union is the outcome space, and so $C$ must be a subset of this union.

Also, because $E,F$ are mutually exclusive (or disjoint), so too are any subsets of each of these, such as $D\cap E,D\cap F$. 

Thus:-

$$\def\P{\operatorname{\sf P}}\begin{align}\P(D)&=\P(D\cap(E\cup F))&&E,F\text{ are exhaustive events}\\&=\P((D\cap E)\cup (D\cap F))&&\text{Distributivity of Intersection over Union}\\&=\P(D\cap E)+\P(D\cap F)&&D\cap E, D\cap F\text{ are mutually exclusive events}\\&=\P(D\mid E)\P(E)+\P(D\mid F)\P(F)&&\text{definition of conditional probability}\end{align}$$