Understanding Bayes Theorem

Question

I'm given a prompt to determine the Probability of a single event.

Known:

P(A | B) = .7

P(A | ~B) = .3

P(B | A) = .6

I need to figure out P(A).

P(A|B) = (P(B | A)*P(A)) / ((P(B | A)*P(A)) + P(B | ~A)*P(~A))

I am a bit lost on how to derive P(B | ~A) and P(~A) from the given known elements.

I don't think P(B | ~A) = (1 - P(B | A)) but even if it is, how do I get P(~A)?

Answer 1

$P(A \cap B)=0.7 P(B)$ and $P(A \cap B)=0.6 P(A)$ from the first and the last equations. Let $P(A)=c$. Then we get $P(B)=\frac {0.6}{0.7}c$. Now $P(A\cap \sim B)=0.3 P(\sim B)=0.3 (1-P(B))=0.3 (1-\frac {0.6}{0.7}c)$ from the second equation. Also $P(A \cap B)=0.6 P(A)$. Add these two equations. You get $c=P(A)=P(A\cap \sim B)+P(A \cap B)=0.3 (1-\frac {0.6}{0.7}c)+0.6 c$. Now solve this equation for $c$.

Answer 2

I think it would be helpful to consider what I regard as the classic conditional probability problem. I offer this problem only as a way of kick-starting the OP's intuition around conditional probability problems.

You are given two weather forecasters : Mr. R and Mr. S.

You are given that Mr. R's forecasts are accurate $80\%$ of the time.

You are given that Mr. S's forecasts are accurate $90\%$ of the time.

One morning, Mr. R says it will rain, and Mr. S says it will not rain

What are the chances of rain?

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It is important to realize that the problem can not be solved without making the following assumptions.

Normally, the forecasts of Mr. R and Mr. S are considered independent events.

Let A denote the event that Mr. R is right and Mr. S is right.
Let B denote the event that Mr. R is right and Mr. S is wrong.
Let C denote the event that Mr. R is wrong and Mr. S is right.
Let D denote the event that Mr. R is wrong and Mr. S is wrong.

$p(A) = 0.8 \times 0.9 = 0.72.$
$p(B) = 0.8 \times 0.1 = 0.08.$
$p(C) = 0.2 \times 0.9 = 0.18.$
$p(D) = 0.2 \times 0.1 = 0.02.$

The problem is (in effect) asking:
Given that either events B or C occurred,
what is the chance that event B occurred.

Answer:
$\frac{p(B)}{p(B) + p(C)} = \frac{.08}{.08 + .18}.$

More formally, in Bayes Theorem parlance:
Let E denote the event that Mr. R and Mr. S disagree.
Then you want $p(B|E)$.

Here, you use the formula that
$p(E) \times p(B|E) = p(B ~\text{and} ~E ~\text{both occur}) = p(B).$
Thus, $p(B|E) = \frac{p(B)}{p(E)} = \frac{.08}{.08 + .18}.$

Answer 3

This can be also visually solved using the basic Venn-Diagram. Let the orange circle represent the set of event A and the green circle represents the set of event B. x is the number of events favourable to only A, y is the number of events favourable both A and B, number of events favourable to only B and w is the number of events not favourable to A or B.

$$P\left(A|B\right)=\frac{y}{y+z}=\frac{7}{10}$$ $$\implies z=\frac{3y}{7}$$

Also, $$P(B|A)=\frac{y}{x+y}=\frac{6}{10}$$ $$\implies x=\frac{2y}{3}$$

Similarly, $$P(A|~B)=\frac{x}{x+w}=\frac{3}{10}$$ $$\implies w=\frac{7x}{3}=\frac{14y}{9}$$

Now, we need to calculate $$P(A)=\frac{\left(x+y\right)}{x+y+z+w}$$

Just substitute the values and you have the answer.

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If there are only 2 sets of events mentioned in the question, this is the easiest way to solve such problems. However, visualizing 3 or more sets of events may be difficult and require considerable practice. So, I would recommend you to stick to the algebraic method, meanwhile having an idea of the Venn-Diagram in the back of your mind if you get stuck.