Probability and Conditional Events

Part (a): Addition rule:

Part (b): Test for independence:

Since $P(A \cap B) = P(A) \times P(B) = 0.28$, events $A$ and $B$ are independent.

Final Answer: (a) $0.82$; (b) Yes, independent because $P(A \cap B) = P(A)P(B) = 0.28$.

Step 1: Apply conditional probability formula:

Step 2: Substitute values:

Final Answer: $0.30$

Step 1: First draw: Total marbles $= 8$. $P(R_1) = \frac{5}{8}$.

Step 2: Second draw (without replacement): Remaining $= 7$, red remaining $= 4$. $P(R_2|R_1) = \frac{4}{7}$.

Step 3: Multiplication rule:

Final Answer: $\frac{5}{14}$

Part (a): By definition of mutually exclusive events, $P(E \cap F) = 0$.

Part (b): For mutually exclusive events:

Final Answer: $P(E \cap F) = 0$; $P(E \cup F) = 0.80$.

Part (a): Total probability axiom:
All four disjoint regions in the universal set must sum to 1: $$0.3 + x + 0.2 + 0.25 = 1 \implies x + 0.75 = 1 \implies x = 0.25$$

Part (b): Calculate $P(A|B)$:
$P(B) = P(A \cap B) + P(A' \cap B) = 0.25 + 0.2 = 0.45$. $$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.25}{0.45} = \frac{5}{9} \approx 0.556$$

Part (c): Independence test:
$P(A) = P(A \cap B') + P(A \cap B) = 0.3 + 0.25 = 0.55$.
Compare $P(A|B) = \frac{5}{9} \approx 0.5556$ with $P(A) = 0.5500$.
Alternatively: $P(A) \times P(B) = 0.55 \times 0.45 = 0.2475 \neq 0.25 = P(A \cap B)$.
Since $P(A \cap B) \neq P(A)P(B)$, events $A$ and $B$ are not independent.

Final Answer: (a) $x = 0.25$; (b) $\frac{5}{9}$; (c) Not independent because $P(A \cap B) = 0.25 \neq 0.2475 = P(A)P(B)$.

Part (a) Branches:
• Saturday Rain ($R_1$): $0.6$
    → Sunday Rain ($R_2|R_1$): $0.7 \implies P(R_1 \cap R_2) = 0.6 \times 0.7 = 0.42$
    → Sunday Dry ($R_2'|R_1$): $0.3 \implies P(R_1 \cap R_2') = 0.6 \times 0.3 = 0.18$
• Saturday Dry ($R_1'$): $0.4$
    → Sunday Rain ($R_2|R_1'$): $0.3 \implies P(R_1' \cap R_2) = 0.4 \times 0.3 = 0.12$
    → Sunday Dry ($R_2'|R_1'$): $0.7 \implies P(R_1' \cap R_2') = 0.4 \times 0.7 = 0.28$

Part (b): Rains on at least one day:
$$P(\text{At least one}) = 1 - P(\text{Dry both}) = 1 - 0.28 = 0.72$$

Part (c): Conditional $P(R_1|R_2)$:
Total probability of rain on Sunday: $$P(R_2) = P(R_1 \cap R_2) + P(R_1' \cap R_2) = 0.42 + 0.12 = 0.54$$ $$P(R_1|R_2) = \frac{P(R_1 \cap R_2)}{P(R_2)} = \frac{0.42}{0.54} = \frac{7}{9} \approx 0.778$$

Final Answer: (b) $0.72$; (c) $\frac{7}{9} \approx 0.778$.

Part (a): Set up equation:
Since $A$ and $B$ are independent, $P(A \cap B) = P(A) \times P(B)$: $$2x(x + 0.1) = 0.12 \implies 2x^2 + 0.2x - 0.12 = 0 \quad \blacksquare$$

Part (b): Solve the quadratic:
Multiply by 50 to clear decimals: $100x^2 + 10x - 6 = 0 \implies 50x^2 + 5x - 3 = 0$.
Factor or use quadratic formula: $$x = \frac{-0.2 \pm \sqrt{0.2^2 - 4(2)(-0.12)}}{2(2)} = \frac{-0.2 \pm \sqrt{0.04 + 0.96}}{4} = \frac{-0.2 \pm \sqrt{1.0}}{4} = \frac{-0.2 \pm 1}{4}$$ Positive root: $x = \frac{0.8}{4} = 0.2$. (Negative root $x = -0.3$ is rejected because probability cannot be negative).

Part (c): Calculate $P(A \cup B)$:
$P(A) = 2(0.2) = 0.4$, $P(B) = 0.2 + 0.1 = 0.3$. $$P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.4 + 0.3 - 0.12 = 0.58$$

Final Answer: (b) $x = 0.2$; (c) $P(A \cup B) = 0.58$.

Mathematical Proof:
Let $P(A) > 0$ and $P(B) > 0$.
1. If $A$ and $B$ are mutually exclusive, then $P(A \cap B) = 0$.
2. If $A$ and $B$ are independent, then $P(A \cap B) = P(A) \times P(B)$.
3. For both conditions to hold simultaneously: $$P(A) \times P(B) = 0$$ This requires either $P(A) = 0$ or $P(B) = 0$, which contradicts the premise that both events have non-zero probabilities.
Conclusion: Two non-impossible events can NEVER be both mutually exclusive and independent.

Final Answer: No; if mutually exclusive, $P(A \cap B) = 0$, but if independent with non-zero probabilities, $P(A \cap B) = P(A)P(B) > 0$.

Part (a): Total probability of positive test ($T^+$):
$P(D) = 0.01 \implies P(D') = 0.99$.
$$P(T^+|D) = 0.98 \implies P(D \cap T^+) = 0.01 \times 0.98 = 0.0098$$ $$P(T^+|D') = 0.05 \implies P(D' \cap T^+) = 0.99 \times 0.05 = 0.0495$$ $$P(T^+) = 0.0098 + 0.0495 = 0.0593$$

Part (b): Reverse conditional probability $P(D|T^+)$:
$$P(D|T^+) = \frac{P(D \cap T^+)}{P(T^+)} = \frac{0.0098}{0.0593} = \frac{98}{593} \approx 0.16526 \approx 16.5\%$$

Part (c): Reflection on counter-intuition:
Even though the test is $98\%$ sensitive, an individual with a positive test has only about a $16.5\%$ chance of actually having the disease. Because the disease is so rare ($1\%$), the absolute number of false positives ($0.0495$) significantly outnumbers the true positives ($0.0098$).

Final Answer: (a) $0.0593$; (b) $\frac{98}{593} \approx 0.165$ ($16.5\%$); (c) The false positive pool dwarfs the true positive pool due to the low disease base rate.

Part (a): Probability of green ball:
Die roll determines box:
$P(B_1) = P(\{1, 2\}) = \frac{2}{6} = \frac{1}{3}$.
$P(B_2) = P(\{3, 4, 5, 6\}) = \frac{4}{6} = \frac{2}{3}$.
In Box 1: $P(G|B_1) = \frac{4}{6} = \frac{2}{3}$.
In Box 2: $P(G|B_2) = \frac{3}{8}$.
$$P(G) = P(B_1 \cap G) + P(B_2 \cap G) = \left(\frac{1}{3} \times \frac{2}{3}\right) + \left(\frac{2}{3} \times \frac{3}{8}\right) = \frac{2}{9} + \frac{6}{24} = \frac{2}{9} + \frac{1}{4} = \frac{8 + 9}{36} = \frac{17}{36} \approx 0.472$$

Part (b): Conditional $P(B_1|G)$:
$$P(B_1|G) = \frac{P(B_1 \cap G)}{P(G)} = \frac{2/9}{17/36} = \frac{2}{9} \times \frac{36}{17} = \frac{8}{17} \approx 0.471$$

Part (c): Second green ball from same box:
Using conditional probabilities given which box was selected:
If from Box 1: Remaining balls $= 3$ green, $2$ yellow (total 5) $\implies P(G_2|B_1 \cap G_1) = \frac{3}{5}$.
If from Box 2: Remaining balls $= 2$ green, $5$ yellow (total 7) $\implies P(G_2|B_2 \cap G_1) = \frac{2}{7}$.
Using Bayes' theorem probabilities of the box given $G_1$ (from (b), $P(B_1|G_1) = \frac{8}{17}$ and $P(B_2|G_1) = \frac{9}{17}$):
$$P(G_2|G_1) = \left(\frac{8}{17} \times \frac{3}{5}\right) + \left(\frac{9}{17} \times \frac{2}{7}\right) = \frac{24}{85} + \frac{18}{119} = \frac{168 + 90}{595} = \frac{258}{595} \approx 0.4336 \approx 0.434$$

Final Answer: (a) $\frac{17}{36}$; (b) $\frac{8}{17}$; (c) $\frac{258}{595} \approx 0.434$.

(a) Intersection:
Students studying at least one subject $= 80 - 15 = 65$. [M1]
$n(C \cap B) = n(C) + n(B) - n(C \cup B) = 45 + 40 - 65 = 20$. [A1]

(b) Conditional probability:
$P(C|B) = \frac{n(C \cap B)}{n(B)} = \frac{20}{40} = 0.5$. [M1 A1]

(c) Independence test:
$P(C) = \frac{45}{80} = 0.5625$.
Since $P(C|B) = 0.50 \neq 0.5625 = P(C)$, the events are not independent. [M1 R1]

(a) Event A:
Total sample space $= 6 \times 6 = 36$.
$A = \{(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)\}$. [A1]
$P(A) = \frac{6}{36} = \frac{1}{6}$. [A1]

(b) Event B and intersection:
$B$ contains all pairs with at least one 6: $(6, 1), (6, 2), \dots, (6, 6)$ ($6$ pairs) plus $(1, 6), (2, 6), \dots, (5, 6)$ ($5$ pairs) $\implies n(B) = 11$.
$P(B) = \frac{11}{36}$. [A1]
$A \cap B = \{(1, 6), (6, 1)\} \implies n(A \cap B) = 2$.
$P(A \cap B) = \frac{2}{36} = \frac{1}{18}$. [A1]

(c) Independence verification:
Compute $P(A) \times P(B) = \frac{1}{6} \times \frac{11}{36} = \frac{11}{216}$. [M1]
Compare with $P(A \cap B) = \frac{1}{18} = \frac{12}{216}$.
Since $\frac{12}{216} \neq \frac{11}{216}$, $P(A \cap B) \neq P(A)P(B)$. [M1]
Therefore, events $A$ and $B$ are not independent. [R1]