# “Incremental” Bayes Rule given another similar test

Let $T$ denote the kit showing positive and $C$ denoting the person being a carrier.

$P(C) = 0.02, P(T|C) = 0.998, P(\neg T | \neg C) = 0.996.$ I've already found $P(\neg C|T) = 0.164$.

The question then asks given that the patient used another kit and is tested positive again, what is the updated likelihood that he is not a carrier.

I then did this: $P(\neg C | T_1 \cap T_2) = \frac{P(\neg C)}{P(T_1 \cap T_2) } P(T_1\cap T_2 | \neg C) = \frac{P(\neg C)}{P(T_1 \cap T_2) } P(T_1 |\neg C)P(T_2|\neg C)$ which was the wrong working.

May I know what went wrong with my understanding?

Thank you.

Your solution so far is actually not wrong. If you go down that path, you can continue as:

1) first, note that $P(T_1|C) = P(T_2|C)$.

2) the denominator $P(T1, T2)$ is equivalent of $P(T1, T2, C) + P(T1, T2, !C)$. Using the chain rule and the fact that T1 and T2 are conditionally independent of each other given C or !C, we can re-write this as $P(T1|C)P(T2|C)P(C) + P(T1|!C)P(T2|!C)P(!C)$

3) altogether, right now the formula is the following, where I use T instead of T1 and T2. $$\frac{P(T|!C)^2P(!C)}{P(T|C)^2P(C) + P(T|!C)^2P(!C)}$$

4) You can sub in the values: $$\frac{0.004^2* 0.98}{0.004^2*0.98+0.998^2*0.02}=0.0008$$

However, as you mentioned, there is an incremental Bayes rule which can take advantage of your solution from the previous part. The formula is: $$P(h|x1, x2)= \frac{P(x2|h)P(h|x1)}{\sum_h P(x2|h)P(h|x1)}$$

Using this formula, you can directly plug in the value of $P(h|x1)$ you had previously found, which was 0.164. It just makes the calculations more simple compared to your original train of thought.