# Prove $P(a \mid (b \text{ and } a)) =1$

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?

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## 2 Answers

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 $p(a|ab)=1$.

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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 $B$.

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 $P(B)$.

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