For a certain probability experiment, the probability that event $A$ will occur is $\displaystyle\frac12$ and the probability that even $B$ will occur is $\displaystyle\frac13$. Which of the following values could be the probability that the event $A \cup B$ (that is, the event $A$ or $B$, or both) will occur?
Indicate all such values.
A) $\displaystyle\frac13$
B) $\displaystyle\frac12$
C) $\displaystyle\frac34$
Answer is $B$ and $C$.
The probability that the event $A \cup B$ (that is, the event $A$ or $B$, or both) will occur lies in the interval 1/2 to 5/6.
How? Please share how to determine the interval for the probability that the event $A \cup B$ (that is, the event $A$ or $B$, or both) will occur.
$$P(A \cup B)=P(A)+P(B)-P(A \cap B)$$
As David Mitra Said you have a sum and a negative term, naturally if $P(A \cap B)=0$ the expression is the largest possible and if $P(A \cap B)=P(A)$ or $P(A \cap B)=P(B)$ the expression is smallest (which is the case when $B \subset A$ or $A \subset B$).
Largest case $P(A \cap B)=0$:
$$P(A \cup B)=P(A)+P(B)-P(A \cap B)=P(A)+P(B)$$
Smallest case one $P(A \cap B)=P(A)$
$$P(A \cup B)=P(A)+P(B)-P(A \cap B)=P(A)+P(B)-P(A)=P(B)$$
Smallest case two $P(A \cap B)=P(B)$
$$P(A \cup B)=P(A)+P(B)-P(A \cap B)=P(A)+P(B)-P(B)=P(A)$$
I let you replace and see which one is the smallest.