1. **Problem statement:** Given probabilities in a Venn diagram with events A, B, C, D and some intersections, find mutually exclusive events, values of unknown probabilities, conditional probabilities, and set expressions. 2. **Mutually exclusive events:** Events that cannot happen together have intersection probability 0. 3. **Given:** - $P(A \cap B) = 0.05$ - $P(B \cap C) = 0.27$ - $P(C \text{ only}) = 0.25$ - $P(D) = 0.08$ (inside C, no intersections with others) - $P(B \text{ only}) = p$ - $P(A \text{ only}) = q$ - $P(\text{outside A}) = r$ 4. **(a) Mutually exclusive events:** - Since $D$ is inside $C$ and does not intersect others, $D$ and $A$ are mutually exclusive. - Also, $D$ and $B$ are mutually exclusive. - So one pair is $(D, A)$. 5. **(b) Find $p$ given $B$ and $C$ are independent:** - Independence means $P(B \cap C) = P(B)P(C)$. - $P(B) = P(B \text{ only}) + P(A \cap B) + P(B \cap C) = p + 0.05 + 0.27 = p + 0.32$. - $P(C) = P(C \text{ only}) + P(B \cap C) + P(D) = 0.25 + 0.27 + 0.08 = 0.6$. - Using independence: $0.27 = (p + 0.32)(0.6)$. - Solve for $p$: $$0.27 = 0.6p + 0.192 \Rightarrow 0.6p = 0.078 \Rightarrow p = \frac{0.078}{0.6} = 0.13$$ 6. **(c) Greatest possible value of $P(A | B')$:** - $P(A | B') = \frac{P(A \cap B')}{P(B')} = \frac{P(A) - P(A \cap B)}{1 - P(B)}$. - $P(A) = q + 0.05 + \text{(intersection with C if any, not given, assume 0)} = q + 0.05$. - $P(B) = p + 0.05 + 0.27 = 0.13 + 0.05 + 0.27 = 0.45$. - To maximize $P(A | B')$, maximize numerator $P(A) - 0.05 = q$ and minimize denominator $1 - 0.45 = 0.55$. - Since total probability is 1, sum all parts: $$q + 0.05 + p + 0.27 + 0.25 + 0.08 + r = 1$$ $$q + 0.05 + 0.13 + 0.27 + 0.25 + 0.08 + r = 1$$ $$q + r + 0.78 = 1 \Rightarrow q + r = 0.22$$ - $r$ is outside $A$, so to maximize $q$, set $r=0$. - Then $q=0.22$. - So: $$P(A | B') = \frac{q}{1 - P(B)} = \frac{0.22}{0.55} = 0.4$$ 7. **(d) Given $P(B | A') = 0.5$, find $q$ and $r$:** - $P(B | A') = \frac{P(B \cap A')}{P(A')} = 0.5$. - $P(A') = 1 - P(A) = 1 - (q + 0.05)$. - $P(B \cap A') = P(B) - P(A \cap B) = 0.45 - 0.05 = 0.4$. - So: $$0.5 = \frac{0.4}{1 - (q + 0.05)} \Rightarrow 1 - (q + 0.05) = \frac{0.4}{0.5} = 0.8$$ $$q + 0.05 = 0.2 \Rightarrow q = 0.15$$ - Using total probability: $$q + 0.05 + p + 0.27 + 0.25 + 0.08 + r = 1$$ $$0.15 + 0.05 + 0.13 + 0.27 + 0.25 + 0.08 + r = 1$$ $$0.93 + r = 1 \Rightarrow r = 0.07$$ 8. **(e) Find $P([A \cup B]' \cap C)$:** - $[A \cup B]'$ is complement of $A \cup B$. - So $P([A \cup B]' \cap C) = P(C) - P((A \cup B) \cap C)$. - $(A \cup B) \cap C = (A \cap C) \cup (B \cap C)$. - No $A \cap C$ given, assume 0. - $P(B \cap C) = 0.27$. - $P(C) = 0.25 + 0.27 + 0.08 = 0.6$. - So: $$P([A \cup B]' \cap C) = 0.6 - 0.27 = 0.33$$ 9. **(f) Expression for event with probability $p$:** - $p$ is $P(B \text{ only})$. - Set notation: $B \cap A' \cap C'$. **Final answers:** - (a) Mutually exclusive pair: $(D, A)$ - (b) $p = 0.13$ - (c) Greatest $P(A | B') = 0.4$ - (d) $q = 0.15$, $r = 0.07$ - (e) $P([A \cup B]' \cap C) = 0.33$ - (f) Event with probability $p$ is $B \cap A' \cap C'$