# Thread: A problem come from "Introduction to Probability Models"

1. ## A problem come from "Introduction to Probability Models"

" If the occurrence of B makes A more likely, does the occurrence of A make B more likely?

I guess yes. But I could not figure out the reason. Thanks~~

2. Hello, shiningstarpxx!

An interesting question . . .

If the occurrence of $B$ makes $A$ more likely,
does the occurrence of $A$ make $B$ more likely?
The answer is "Yes".
We need Bayes' Theorem: . $P(X\,|\,Y) \;=\;\frac{P(X \wedge\,Y)}{P(Y)}$

We are told that, if $B$ happens, $A$ is more likely to happen.
. . That is, the probability of $A$, given $B$, is greater than the probability of $A$.

In symbols: . $P(A\,|\,B) \:> \:P(A)$

This means: . $\frac{P(A \wedge B)}{P(B)} \:> \:P(A) \quad\Rightarrow\quad P(A\wedge B) \:>\:P(A)\!\cdot\!P(B)$

Hence, we have: . $\frac{P(B \wedge A)}{P(A)} \:>\:P(B) \quad\Rightarrow\quad P(B\,|\,A) \:>\:P(B)$

Therefore, the probability of $B$, given $A$, is greater than the probability of $B.$
. . The occurence of $A$ makes $B$ more likely.

3. Originally Posted by shiningstarpxx
" If the occurrence of B makes A more likely, does the occurrence of A make B more likely?

I guess yes. But I could not figure out the reason. Thanks~~
What you're asking is the following:

Does Pr(A | B) > Pr(A) => Pr(B | A) > Pr(B) ?

The answer is yes. Consider:

$\Pr(A | B) \, \Pr(B) = \Pr(B | A) \, \Pr(A) \Rightarrow \frac{\Pr(A | B)}{\Pr(A)} = \frac{\Pr(B | A)}{\Pr(B)}$.

But it's given that $\Pr(A | B) > \Pr(A) \Rightarrow \frac{\Pr(A | B)}{\Pr(A)} > 1$.

Therefore $\frac{\Pr(B | A)}{\Pr(B)} > 1$ and the implication follows.

Edit: Too fast for me this time, Soroban. But I like to think our replies ...... complement each other lol!
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