Use overlap bounds before adding the conditional information

The overlap cannot exceed the smaller marginal. Since $5/8<2/3$, the maximum is $5/8$, attained when A is contained in B, as in the source’s nested-circle construction.

Inclusion–exclusion gives $P(A\cap B)=P(A)+P(B)-P(A\cup B)\ge5/8+2/3-1=7/24$. Equality is attained when $A\cup B$ fills the sample space. The source’s overlap-strip construction and highlighted explanation illustrate this equality case. These extremes use only the two initial marginals, before the later conditional datum.

$P(B')=1-2/3=1/3$. Thus $3/8=P(A'\cap B')/P(B')$ gives $P(A'\cap B')=(3/8)(1/3)=1/8$.

By De Morgan’s law, $P(A\cup B)=1-P(A'\cap B')=1-1/8=7/8$.

From $7/8=5/8+2/3-P(A\cap B)$, the intersection is $5/12$. Also $P(A)P(B)=(5/8)(2/3)=5/12$. Equality proves independence.

$P(A')=1-5/8=3/8=P(A'\mid B')$. Since $P(B')>0$, the complement events A′ and B′ are independent. Independence is preserved by taking complements, so A and B are independent too.

$\begin{aligned}&P(A\cup B\cup C)\\ & =P(A)+P(B)+P(C)\\ &{}-P(A\cap B)-P(A\cap C)-P(B\cap C)\\ &{}+P(A\cap B\cap C).\end{aligned}$

Substitute the source values:

$\begin{aligned}&\frac{11}{12}\\ & =\frac58+\frac23+\frac38-\frac5{12}-\frac14-\frac14\\ &{}+P(A\cap B\cap C).\end{aligned}$

Hence $P(A\cap B\cap C)=1/6$. The retained three-circle region diagram shows why the triple overlap is restored once after subtracting all pairwise overlaps.