Independent vs mutually exclusive events: Venn diagram exam question

The question

The solution

1. 01Draw two overlapping circles inside a box. The given $0.09$ is $A$ without $B$, the given $0.49$ is $B$ without $A$, and the overlap is $x$.
2. 02Read the two events off the diagram: $P(A) = 0.09 + x$ and $P(B) = 0.49 + x$.
3. 03Independent means the probabilities multiply, so $x = P(A)\,P(B) = (0.09 + x)(0.49 + x)$. Do not confuse this with mutually exclusive, which would be two circles that do not touch. If you set $P(A \cap B) = 0$, the question contradicts itself, because $0.09 \times 0.49 \neq 0$.
4. 04Multiply out to get $x = x^{2} + 0.58x + 0.0441$, then bring everything to one side: $x^{2} - 0.42x + 0.0441 = 0$. The four marks are for reaching this line.
5. 05The discriminant is $\Delta = 0.42^{2} - 4(0.0441) = 0.1764 - 0.1764 = 0$, so there is one repeated root and no second answer to discard. Hence $x = \frac{0.42}{2} = 0.21$.
6. 06Check: $P(A) = 0.3$ and $P(B) = 0.7$, and $0.3 \times 0.7 = 0.21$, so the result is consistent.
7. 07For part (b), use $P(A' \mid B') = \frac{P(A' \cap B')}{P(B')}$. Inside the circles the total is $0.09 + 0.21 + 0.49 = 0.79$, so outside both is $1 - 0.79 = 0.21$, and $P(B') = 1 - 0.7 = 0.3$. This gives $\frac{0.21}{0.3} = 0.7$.
8. 08Shortcut for part (b): if $A$ and $B$ are independent, so are their complements, so $P(A' \mid B') = P(A') = 0.7$. A conditional probability equal to the plain one is independence showing itself, and it is worth stating in the answer.

Answer
x = 0.21 and P(A′ | B′) = 0.21/0.3 = 0.7.