Communities

Writing
Writing
Codidact Meta
Codidact Meta
The Great Outdoors
The Great Outdoors
Photography & Video
Photography & Video
Scientific Speculation
Scientific Speculation
Cooking
Cooking
Electrical Engineering
Electrical Engineering
Judaism
Judaism
Languages & Linguistics
Languages & Linguistics
Software Development
Software Development
Mathematics
Mathematics
Christianity
Christianity
Code Golf
Code Golf
Music
Music
Physics
Physics
Linux Systems
Linux Systems
Power Users
Power Users
Tabletop RPGs
Tabletop RPGs

Dashboard
Notifications
Mark all as read
Q&A

In general, does $\color{forestgreen}{P(A|M)} + \color{red}{P(B|M)} = 1$?

+1
−1

In this question, $\color{forestgreen}{P(A|M)} + \color{red}{P(B|M)} = 1$. But the author's solution didn't annunciate this, and doesn't unfurl how to compute $\color{red}{P(B|M)}$. Can I simply subtract as $\color{red}{P(B|M)} = 1 - \color{forestgreen}{P(A|M)}$? I hanker to avoid calculating $\color{red}{P(B|M)}$ by applying Bayes's Rule a second time.

  1. A crime is committed by one of two suspects, A and B. Initially, there is equal evidence against both of them. In further investigation at the crime scene, it is found that the guilty party had a blood type found in 10% of the population. Suspect A does match this blood type, whereas the blood type of Suspect B is unknown.

(a) Given this new information, what is the probability that A is the guilty party?

(b) Given this new information, what is the probability that B's blood type matches that found at the crime scene?

Solution:

(a) Let M be the event that A's blood type matches the guilty party's and for brevity, write A for "A is guilty" and B for "B is guilty". By Bayes' Rule,

Image alt text

Blitzstein, Introduction to Probability (2019 2 edn), Ch 2, Exercise 25, p 87. p 12 in the publicly downloadable PDF of curbed solutions.

Why does this post require moderator attention?
You might want to add some details to your flag.
Why should this post be closed?

0 comment threads

1 answer

+0
−0

This is a special instance of the more general fact that $P(A)+P(B)=1$ if $A$ and $B$ "partition" the probability space. I'll explain what I mean.

A partition of a set is any way of carving up the set into (1) disjoint and (2) exhaustive subsets. For instance if the set is {1,2,3,4} then a partition of this set consists of the set of {1}, {2,3}, {4}. There are many other partitions, but this is one example. Notice that it is disjoint because no two "cells" of the partition "overlap". Notice also that it's exhaustive because every element of the entire set ({1,2,3,4}) is in one of the cells of the partition.

Now look at your problem. We are told that either A committed the crime or B did. That's effectively a partition over the probability space. I'll avoid talking too much about exactly what the probability space is here, because that may be a long and not extremely important digression right now. But intuitively think of it like this: Disjointness basically corresponds to the fact that A and B cannot both have committed the crime (no overlap in the "events"). Exhaustiveness means that there is no other scenario than these two.

When two events form a partition, it is always true that $P(A)+P(B)=1$.

In this particular case we are conditioning on the event $M$. It is possible to prove that when you condition on some event, what results is another probability space where all the usual rules apply. And in this new probability space, it is still true that events $A$ and $B$ form a partition, so it is still true that $P(A|M)+P(B|M)=1$.


Ok, that answers the question, but there is a different response that you might find even more useful: What are you supposed to know for this problem? You probably should NOT commit to memory that a partition has this property. Rather, you should be able to derive this fact from more basic facts. In particular, when events are disjoint they obey an additive rule. That is to say, when $A$ and $B$ are disjoint (but not necessarily exhaustive and therefore not necessarily a partition) it is still true that $P(A\cup B)=P(A)+P(B)$. That's one thing you should know.

Another thing you should know is that the probability taken over the entire space is 1. So if $A$ and $B$ are exhaustive (but not necessarily disjoint, so not necessarily a partition) it is true that $P(A\cup B)=P(X)=1$ where $X$ is the entire probability space.

Now if you know that $A$ and $B$ form a partition, you can prove that $P(A)+P(B)=1$ through the following short sequence of equations:

$$ P(A)+P(B)=P(A\cup B) = P(X)=1 $$

The first equation is due to disjointness (and therefore obeying the additive rule). The second equation is due to exhaustiveness. And the third is due to the fact that the probability taken over the entire space is always 1.


Yet MORE generally than this, you should know the rule for the probability of a general union. For ANY two events, it is always true that $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. If the two events are disjoint then $A\cap B=\emptyset$ and therefore $P(A\cap B)=P(\emptyset)=0$. In this case, it then becomes clear that disjointness implies $P(A\cup B)=P(A)+P(B)$.

Why does this post require moderator attention?
You might want to add some details to your flag.

0 comment threads

Sign up to answer this question »

This community is part of the Codidact network. We have other communities too — take a look!

You can also join us in chat!

Want to advertise this community? Use our templates!

Like what we're doing? Support us! Donate