Using the formula P(B)=P(A)P(Bmid A)+P(overline…
Using the formula $P(B)=P(A)P(B\mid A)+P(\overline A)P(B\mid\overline A)$, show that if the random events $A$ and $B$ are independent, then the events $\overline A$ and $B$ are also independent.
Explanation of Method
Independence of $A$ and $B$ means $P(B\mid A)=P(B)$ (when $P(A)\gt0$). Put this into the given formula (the total probability formula for the complete system $A,\overline A$) and solve for $P(B\mid\overline A)$: it turns out to equal $P(B)$, which is exactly the independence of $\overline A$ and $B$. We also give a proof without conditional probabilities that covers the edge cases.
Step 1: The formula
$A$ and $\overline A$ form a complete system, so $B=AB+\overline AB$ and
\[ P(B)=P(A)P(B\mid A)+P(\overline A)P(B\mid\overline A). \]
Assume $0\lt P(A)\lt1$, so both conditional probabilities exist.
Step 2: Use independence
$A$ and $B$ independent means $P(B\mid A)=P(B)$. Then
\[ P(B)=P(A)P(B)+P(\overline A)P(B\mid\overline A) \quad\Longrightarrow\quad P(B)\big(1-P(A)\big)=P(\overline A)P(B\mid\overline A). \]
Since $1-P(A)=P(\overline A)\gt0$, divide:
\[ \boxed{P(B\mid\overline A)=P(B),} \]
i.e. $\overline A$ and $B$ are independent: $P(\overline AB)=P(\overline A)P(B)$. $\blacksquare$
Step 3: A proof without conditional probabilities
\[ P(\overline AB)=P(B)-P(AB)=P(B)-P(A)P(B)=\big(1-P(A)\big)P(B)=P(\overline A)P(B). \]
This works also when $P(A)=0$ or $P(A)=1$.
Step 4: Consequences
Applying the result again (to $B$ and then to $\overline A$) shows: if $A$ and $B$ are independent, then so are each of the pairs
\[ (\overline A,B),\qquad (A,\overline B),\qquad (\overline A,\overline B). \]
Meaning: if knowing that $A$ occurred tells nothing about $B$, then knowing that $A$ did not occur tells nothing about $B$ either.
Result
| Key step | $P(B)=P(A)P(B)+P(\overline A)P(B\mid\overline A)\Rightarrow P(B\mid\overline A)=P(B)$ |
| Conclusion | $\overline A$ and $B$ are independent (also $A,\overline B$ and $\overline A,\overline B$) |