# Probability interval givem symmetric difference and the union of two events

The events $$E$$, $$F$$ and $$P$$ satisfy the constraints $$(E \triangle F) = .7$$, $$P (E ∪ F) = .8$$.

• (A.) Calculate $$P (E ∩ F)$$, making sure that is uniquely determined.
• (B.) Determine the range of possible values of $$P (E)$$.
• (C.) There may be independent $$E,F$$ events satisfying the constraints?

So I have done (A.) $$P (E ∩ F) = .1$$ and $$P(E)+P(F)=.9$$

At this point, I think that for (B.) $$P(E)$$ should be $$<.9$$ but I can't find the low limit.

For point (C.) I know that $$P (E ∩ F)=P(E)P(F)$$ if independent events. So I think I have to replace it on $$P (E ∪ F)=P(E)+P(F)-P (E ∩ F)$$ and see if the values ould exist.

For $(B)$, observe that $$P(E) \geq P(E \cap F) = 0.1$$ Also observe that you have over estimated the upper bound since $$P(E) \leq P(E \cup F) = 0.8$$ For $(C)$, if exist, then $$\begin{cases} P(E) \cdot P(F) = 0.1 \\ P(E) + P(F) = 0.9 \end{cases}$$ Solving these equations, we can obtain one solution as $$P(E) = \frac{0.9 + \sqrt{0.41}}{2},\quad P(F) = \frac{0.9 - \sqrt{0.41}}{2} \tag{1}$$ One example achieving these probabilities is a uniform random variable $X$ on $[0, 1]$. Let $E$ be the event that $\{X \leq \frac{0.9 + \sqrt{0.41}}{2}\}$ and $F$ be the event that $\{0.8 \geq X \geq \frac{0.7 + \sqrt{0.41}}{2}\}$. Then the probability of $E$ and $F$ satisfy $(1)$ and the conditions in the question.