Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

So I am supposed to prove that $P(a)$ given $P(b \text{ and } a)$ is $1$

One way of solving this is that with $P(b \text{ and } a )$ as given the sample space has been reduced to that only, and for that sample space the $a$ is there always so answer is $1$.

But when I tried to do that using bayes theorem, I couldn't solve it

$P(b \text{ and } a \mid a)\cdot P(a)$

divided by

$P(b \text{ and } a)$

Can anyone give me the explanation for this?

share|improve this question

2 Answers 2

Bayes formula amounts to $p(x|y)p(y)=p(xy)$ by interchanging the roles of $x$ and $y$. So, here is a formal proof: $p(a|ab)\cdot p(ab)=p(aab)=p(ab)$. If $p(ab)\ne 0$ then you can divide by it to obtain $p(a|ab)=1$.

share|improve this answer

You simply need to come back to the definition of conditional probabilities. Given two events $A,B$, the probability of $A$ given $B$ is defined as $$P(A|B)=\frac{P(A\cap B)}{P(B)}.$$ Of course, as you have certainly seen, it corresponds to the intuitive notion of "conditonal probability": we restrict the sample space to $B$.

enter image description here

Now, apply this to compute $P(A | A\cap B)$.

Bayes Therorem follows from a simple application of this definition. You usually use it when you want to compute $P(A|B)$ knowing $P(B|A)$, $P(A)$ and $P(B)$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.