Given p=P(A) , q=P(B) , r=P(A+B) . Find P(Atriangle B)…

Given $p=P(A)$, $q=P(B)$, $r=P(A+B)$. Find $P(A\triangle B)$, $P(A\overline{B})$ and $P(\overline{AB})$.

Step 1: The probability of $AB$

The addition rule for two events is

\[ P(A+B) = P(A) + P(B) - P(AB). \]

(Adding $P(A)$ and $P(B)$ counts the common part $AB$ twice, so it is subtracted once.) With our notation $r = p + q - P(AB)$, hence

\[ P(AB) = p + q - r. \]

Step 2: $P(A\overline{B})$ — "$A$ occurs, $B$ does not"

$A$ splits into two incompatible pieces: $A = AB + A\overline{B}$. So $P(A) = P(AB) + P(A\overline{B})$ and

\[ P(A\overline{B}) = P(A) - P(AB) = p - (p+q-r) = r - q. \]

\[ \boxed{P(A\overline{B}) = r - q.} \]

Another way to see it: $A+B = A\overline{B} + B$ (incompatible), so $r = P(A\overline{B}) + q$.

Step 3: $P(A\triangle B)$ — "exactly one occurs"

$A\triangle B = A\overline{B}+\overline{A}B$ with incompatible terms. By symmetry with Step 2, $P(\overline{A}B) = r - p$. So

\[ P(A\triangle B) = (r-q) + (r-p) = 2r - p - q. \]

Check another way: "at least one" = "exactly one" + "both", so $P(A\triangle B) = P(A+B) - P(AB) = r - (p+q-r) = 2r-p-q$. ✓

\[ \boxed{P(A\triangle B) = 2r - p - q.} \]

Step 4: $P(\overline{AB})$ — "not both occur"

By the rule for the opposite event, $P(\overline{X}) = 1 - P(X)$:

\[ P(\overline{AB}) = 1 - P(AB) = 1 - p - q + r. \]

\[ \boxed{P(\overline{AB}) = 1 - p - q + r.} \]

Step 5: Numerical example

Let $p = 0.5$, $q = 0.4$, $r = 0.7$. Then

Check: the four pieces add up to $0.3+0.2+0.2+0.3=1$ ✓, and $P(A\overline{B})+P(AB)=0.3+0.2=0.5=p$ ✓.

Step 6: Which values of $p,q,r$ are possible?

All piece probabilities must be between $0$ and $1$. This gives

\[ \max(p,q) \le r \le \min(1,\ p+q). \]

For example $p=0.5$, $q=0.4$, $r=0.3$ is impossible: $A+B$ cannot be less likely than $A$ alone.