Why you need to divide to calc conditional probability

Question

The definition of P(A|B) = P(A AND B) / P(B), and it basically translates to "if B happens, what is the probability to A to happen", if this is correct then why simple P(A|B) = P(A AND B) isn't enough?

Thanks!

Answer 1

You might want to multiply through by $P(B)$ and try to understand

$$ P(A, B) = P(A|B) P(B). $$

This says that the probability that $A$ and $B$ both happen can be computed by first taking the probability that $B$ happens, then multiply by the probability that $A$ will happen given that $B$ happens.

Answer 2

Think about this: if $B$ is very unlikely but when it happens $A$ becomes likely then $P(A \text{ and } B)$ is small while $P(A|B)$ is large.

I am extremely unlikely to win the lottery jackpot this weekend ($B$) but if I do I am likely to become a millionaire ($A$), so the probability I win the lottery jackpot and then become a millionaire $P(A \text{ and } B)$ is small, but the probability I become a millionaire if I win the lottery jackpot $P(A|B)$ is high.

Answer 3

It follows from the following insight:

The probability that A and B happen is the probability that B happens, multiplied by the probability that A happens given that B has already happened.

or in symbols

$$P(A, B) = P(A|B)P(B)$$

This rearranges into

$$P(A|B) = \frac{P(A, B)}{P(B)}$$

Answer 4

Suppose that I reach into my pocket, pick a coin at random, and toss it. $A$ is the event the coin comes up heads, and $B$ is the event the coin has heads on both sides. Clearly $P(A\mid B)=1$.

As it happens, there are $10$ coins in my pocket, and only one of them has heads on both sides. When I reach blindly into my pocket and pull out a coin, there’s only one chance in $10$ that I get the two-headed coin, so $P(A~\mathbf{and}~B)$ can be at most $\frac1{10}$, the probability of $B$ alone. In this example $P(A\mid B)$ clearly cannot be equal to $P(A~\mathbf{and}~B)$, and it’s easy to come up with many similar examples. Such examples don’t show what the right calculation is, but they do show clearly that in general $P(A\mid B)\ne P(A~\mathbf{and}~B)$; other answers have explained the correct formula.