1. Express the probability of $B'$ as the sum of its mutually exclusive components involving $A$.

$\mathrm{P}(B') = \mathrm{P}(A \cap B') + \mathrm{P}(A' \cap B')$

2. Substitute the given values to solve for $\mathrm{P}(A \cap B')$.

$0.51 = \mathrm{P}(A \cap B') + 0.15$
$\mathrm{P}(A \cap B') = 0.51 - 0.15 = 0.36$

1. Calculate $\mathrm{P}(A \cap B')$ from the given probabilities (which corresponds to solving part (a)).

From the definition of set differences in probability,
$\mathrm{P}(A \cap B') = \mathrm{P}(B') - \mathrm{P}(A' \cap B')$
$= 0.51 - 0.15 = 0.36$

2. Relate the conditional probability $\mathrm{P}(B \mid A)$ to $\mathrm{P}(A)$ and $\mathrm{P}(A \cap B')$.

Using the conditional probability formula,
$\mathrm{P}(B \mid A) = \frac{\mathrm{P}(A \cap B)}{\mathrm{P}(A)}$
Since $A$ is the union of mutually exclusive events $(A \cap B)$ and $(A \cap B')$, we have $\mathrm{P}(A \cap B) = \mathrm{P}(A) - \mathrm{P}(A \cap B')$.
$\mathrm{P}(B \mid A) = \frac{\mathrm{P}(A) - \mathrm{P}(A \cap B')}{\mathrm{P}(A)}$

3. Substitute the known values to form an equation and solve for $\mathrm{P}(A)$.

$\frac{11}{29} = \frac{\mathrm{P}(A) - 0.36}{\mathrm{P}(A)}$
Multiplying both sides by $29\mathrm{P}(A)$:
$11\mathrm{P}(A) = 29(\mathrm{P}(A) - 0.36)$
$11\mathrm{P}(A) = 29\mathrm{P}(A) - 10.44$
$18\mathrm{P}(A) = 10.44$
$\mathrm{P}(A) = 0.58$

1. Establish the bounds for $C$ based on its mutual exclusivity with $B$.

From parts (a) and (b), we know:
$\mathrm{P}(A \cap B') = 0.36$
$\mathrm{P}(A) = 0.58 \implies \mathrm{P}(A') = 1 - 0.58 = 0.42$
We are given $\mathrm{P}(A' \cap B') = 0.15$.

Since $B$ and $C$ are mutually exclusive, $\mathrm{P}(B \cap C) = 0$. This means $C$ must lie entirely within $B'$, so $C$ is partitioned between $(A \cap B')$ and $(A' \cap B')$.
This restricts the intersections of $C$:
$\mathrm{P}(A \cap C) \le \mathrm{P}(A \cap B') \implies \mathrm{P}(A \cap C) \le 0.36$
$\mathrm{P}(A' \cap C) \le \mathrm{P}(A' \cap B') \implies \mathrm{P}(A' \cap C) \le 0.15$

2. Express the intersection probabilities using the independence of $A$ and $C$.

Since $A$ and $C$ are independent, it follows that $A'$ and $C$ are also independent. We can rewrite the intersections in terms of $\mathrm{P}(C)$:
$\mathrm{P}(A \cap C) = \mathrm{P}(A)\mathrm{P}(C) = 0.58\mathrm{P}(C)$
$\mathrm{P}(A' \cap C) = \mathrm{P}(A')\mathrm{P}(C) = 0.42\mathrm{P}(C)$

3. Solve the inequalities to determine the overall maximum value of $\mathrm{P}(C)$.

Substitute the expressions from Step 2 into the boundary limits from Step 1:

1. $0.58\mathrm{P}(C) \le 0.36 \implies \mathrm{P}(C) \le \frac{36}{58} = \frac{18}{29} \approx 0.621$
2. $0.42\mathrm{P}(C) \le 0.15 \implies \mathrm{P}(C) \le \frac{15}{42} = \frac{5}{14} \approx 0.357$

For $C$ to exist within both valid regions, $\mathrm{P}(C)$ must satisfy both inequalities, meaning it cannot exceed the smaller of the two bounds.

$\text{Maximum } \mathrm{P}(C) = \frac{5}{14}$