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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.

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1 Answer

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$$P(A \cup B)=P(A)+P(B)-P(A \cap B)$$

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.

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