# B Bayes’ Theorem

Brain teaser: A heritable disease occurs randomly in 10% of the population. If someone has the disease, it is passed on to their children with probability 50%. A mother has 1 healthy child. Given this, what’s the conditional probability that the mother has the disease?

• Is the answer 10%? Less? More? How do we quantify it?
• Let $$M$$ be the event that the mother has the disease.
• Let $$C$$ be the event that the child has the disease.
• We want $$P(M\mid \textrm{not } C)$$. We have $$P(M)=0.1$$ and $$P(\textrm{not }C\mid M)=0.5$$.

Solution:

$P(M \mid \textrm{not } C) = \frac{P(\textrm{not } C \mid M)P(M)}{P(\textrm{not } C)}$

So we still need $$P(\textrm{not } C)$$. This could happen in 2 ways (“law of total probability”)

$P(\textrm{not } C)=P(\textrm{not } C \mid M)P(M) + P(\textrm{not } C \mid \textrm{not } M)P(\textrm{not } M)$

We know $$P(\textrm{not } M)=1-P(M)=0.9$$. And we assume $$P( C \mid \textrm{not } M)=0.1$$ because the child can randomly get the disease like anyone else, so then $$P(\textrm{not } C \mid \textrm{not } M)=1-P( C \mid \textrm{not } M)=0.9$$. Finally, then, we’re left with:

$P(M \mid \textrm{not } C) = \frac{0.5 \times 0.1}{0.5\times 0.1 + 0.9 \times 0.9}$

Sanity check: this is less than 10%. That’s what our intuition told us.

• We can get what we need using Bayes’ Theorem.
• We’ve seen above that, for events $$A$$ and $$B$$, $$P(A,B)=P(A\mid B)P(B)$$.
• We can also write this as $$P(A,B)=P(B\mid A)P(A)$$.
• Since these are equal, we get the famous Bayes’ theorem:

$P(A\mid B)=\frac{P(B\mid A)P(A)}{P(B)}$

(0.5*0.1)/(0.5*0.1+0.9*0.9)
## [1] 0.05813953

Follow-up question: What’s the conditional probability that the next child will have the disease?

Suppose a drug test produces a positive result with probability 0.99 for drug users, $$P(T = 1\mid D = 1) = 0.99$$. It also produces a negative result with probability $$0.99$$ for non-drug users, $$P(T = 0\mid D = 0) = 0.99$$. The probability that a random person uses the drug is $$0.001$$, so $$P(D = 1) = 0.001$$. What is the probability that a random person who tests positive is not actually a user, $$P(D = 0 \mid T = 1)$$?