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If $P(A) = 1/3$ and $P(B^{\complement}) = 1/4$, then, can $A$ and $B$ be mutually exclusive?
I already know that for $A$ and $B$ to be mutually exclusive, $A \cap B = \varnothing$ and $P( A \cup B ) = P (A) + P(B)$.
I just can't proceed further than this to prove if $A$ and $B$ are mutually exclusive though as I feel like this is not enough information to determine that.
please help
Thank You
Since $P(B^{\complement})$ (which is $B$ complement) is equal to $\frac{1}{4}$, $P(B) = \frac{3}{4}$.
Now $P(A)+P(B)-P(A \cap B)$ must sum to less than $1$. However, $P(A)+P(B)$ is $\frac{1}{3} + \frac{3}{4} = \frac{13}{12}$. The minimum probability of $P(A \cap B)$ is therefore $\frac{1}{12}$. What can you conclude, given your first condition? (which is for $A,B$ to be mutually exclusive, $A \cap B = \varnothing$.)
Hint:
If they are mutually exclusive then $P(A)+P(B)=P(A\cup B)\leq1$.
Check whether that necessary condition is satisfied and draw conclusions.
$P(B^{\complement}) = 1/4$ gives that $P(B) = 1 - 1/4 = 3/4$
$P(A) = 1/3$
This gives
$P(A) + P(B) = 13/12$
You stated that $P(A \cup B) = P(A) + P(B)$ is a necessary condition for A and B being mutually exclusive.
If this were the case, the event $(A \cup B)$ would have probability 13/12 which is larger than 1.
But no event can have a probability larger than 1. Therefore, it cannot be true that $P(A \cup B) = P(A) + P(B)$.
Since this was a necessary condition for them being mutually exclusive, that cannot be true either.
