Let $A, B$ be two events with $P(A) = 0.2$ and $P(B) = 0.4$. Then which of the following cannot be the possible value of $P(A \cup B)$?
A) $0.3$
B) $0.4$
C) $0.5$
D) $0.6$
I understand that $0.6$ is possible, because if the events are mutually exclusive then $P (AB) = 0$. I also understand that $0.5$ is possible (or $0.52$) if the events are independent because $P(AB) = P(A) \times P(B) = 0.08$, and $.6-.08= .52$.
The correct answer is A) $0.3$. How do we know that it cannot be B) $0.4$?
Thanks in advance.
Best Answer
The first answer must be wrong. If there's a $40\%$ chance of event $B$ happening, there can't only be a $30 \%$ chance of either $A$ or $B$ happening. The event "$A$ or $B$" is at least as likely as "$B$."
You also ask why $0.4$ is a possible probability for $A \cup B$. Imagine that you are picking an integer at random between $1$ and $5$. Let $A$ be the event "I pick $2$", and let $B$ be the event "I pick an even number."