Game Theory and Strategy · Quiz 2 · May 2025

**Step 1: Determine Player 2's dominant action in each state of nature \(\theta \in \{0, 1\}\).** Player 2 knows the true state \(\theta\): - **When \(\theta = 0\)**: - If Player 1 plays \(U\): Player 2 gets \(-1\) from \(L\) and \(1\) from \(R\) (\(R > L\)). - If Player 1 plays \(D\): Player 2 gets \(1\) from \(L\) and \(-1\) from \(R\) (\(L > R\)). - **When \(\theta = 1\)**: - If Player 1 plays \(U\): Player 2 gets \(1\) from \(L\) and \(-1\) from \(R\) (\(L > R\)). - If Player 1 plays \(D\): Player 2 gets \(1\) from \(L\) and \(-1\) from \(R\) (\(L > R\)). Notice that in state \(\theta = 1\), **\(L\) strictly dominates \(R\)** for Player 2! Thus, in any Bayesian Nash Equilibrium, Player 2 of type \(\theta = 1\) **must play \(L\)**. **Step 2: Find Player 1's best response and Player 2's response at \(\theta = 0\).** Given that Player 2 plays \(L\) when \(\theta = 1\): - If Player 2 plays \(R\) when \(\theta = 0\), Player 1's expected payoff is: - From \(U\): \(\frac{1}{2}(u_1(U, R, \theta=0)) + \frac{1}{2}(u_1(U, L, \theta=1)) = \frac{1}{2}(-1) + \frac{1}{2}(1) = 0\). - From \(D\): \(\frac{1}{2}(u_1(D, R, \theta=0)) + \frac{1}{2}(u_1(D, L, \theta=1)) = \frac{1}{2}(1) + \frac{1}{2}(-1) = 0\). - And for Player 2 at \(\theta = 0\) to play \(R\), Player 1 must play \(U\) (since when Player 1 plays \(U\), Player 2 at \(\theta = 0\) gets \(1 > -1\) by playing \(R\)!). Thus, the pure-strategy Bayesian Nash Equilibrium is **(Player 1 plays \(U\); Player 2 plays \(R\) if \(\theta = 0\) and \(L\) if \(\theta = 1\))**. **Answer: A (U)** A: **Correct:** In the pure-strategy BNE, Player 1 plays \(U\), while Player 2 plays \(R\) when \(\theta = 0\) and \(L\) when \(\theta = 1\). B: **Incorrect:** If Player 1 played \(D\), Player 2 at \(\theta = 0\) would play \(L\), in which case Player 1 would get \(-1\) in both states and deviate to \(U\) (to get \(+1\) in both states).

From Q#120, under the Bayesian Nash Equilibrium where Player 1 plays \(U\), Player 2 in state \(\theta = 0\) compares payoffs along row \(U\) of the \(\theta = 0\) table: - Playing \(L\) gives Player 2 a payoff of \(-1\). - Playing \(R\) gives Player 2 a payoff of \(+1\). Since \(1 > -1\), Player 2 who knows \(\theta = 0\) will play **R**. **Answer: B (R)** A: **Incorrect:** When \(\theta = 0\) and Player 1 plays \(U\), playing \(L\) yields \(-1 < 1\). B: **Correct:** When \(\theta = 0\) and Player 1 plays \(U\), Player 2 maximizes payoff by playing \(R\) (earning \(1 > -1\)).

In state \(\theta = 1\), let us inspect Player 2's payoffs: - If Player 1 plays \(U\): \(L\) gives \(1\), while \(R\) gives \(-1\). - If Player 1 plays \(D\): \(L\) gives \(1\), while \(R\) gives \(-1\). Thus, \(L\) is a **strictly dominant strategy** for Player 2 when \(\theta = 1\), so Player 2 will play **L**. **Answer: A (L)** A: **Correct:** When \(\theta = 1\), \(L\) yields \(+1\) regardless of Player 1's action, strictly dominating \(R\) (\(-1\)). B: **Incorrect:** \(R\) is strictly dominated by \(L\) when \(\theta = 1\).

**Step 1: Write Player 2's payoff function \(v_2(a_1, a_2)\) when \(a_1 = 2\).** Given \(v_2(a_1, a_2) = a_2 - 2a_1 + a_1 a_2 - a_2^2\), substitute \(a_1 = 2\): \[v_2(2, a_2) = a_2 - 2(2) + 2a_2 - a_2^2 = 3a_2 - 4 - a_2^2\] **Step 2: Maximize \(v_2(2, a_2)\) with respect to \(a_2 \in [0, 2]\).** \[\frac{\partial v_2}{\partial a_2} = 3 - 2a_2 = 0 \implies a_2^* = 1.5\] **Answer: A (1.5)** A: **Correct:** Setting \(\frac{\partial v_2}{\partial a_2} = 1 + a_1 - 2a_2 = 0\) with \(a_1 = 2\) gives \(a_2 = \frac{1 + 2}{2} = 1.5\). B: **Incorrect:** \(a_2 = 1\) is the best response when \(a_1 = 1\), not when \(a_1 = 2\). C: **Incorrect:** \(a_2 = 1.25\) is the best response when \(a_1 = 1.5\). D: **Incorrect:** \(a_2 = 0\) yields \(-4\), whereas \(a_2 = 1.5\) yields \(-1.75\).

**Step 1: Find the general best-response function of party \(i\).** Differentiating \(v_i(a_i, a_j) = a_i - 2a_j + a_i a_j - a_i^2\) with respect to \(a_i\): \[\frac{\partial v_i}{\partial a_i} = 1 + a_j - 2a_i = 0 \implies a_i^*(a_j) = \frac{1 + a_j}{2}\] **Step 2: Solve for the symmetric Nash equilibrium \(a_1^* = a_2^* = a^*\).** \[a^* = \frac{1 + a^*}{2} \implies 2a^* = 1 + a^* \implies a^* = 1\] **Answer: B (1)** A: **Incorrect:** At \(a_1 = a_2 = 1.5\), each player's best response is \((1 + 1.5)/2 = 1.25 \ne 1.5\). B: **Correct:** Solving \(a^* = (1 + a^*)/2\) gives the unique symmetric Nash equilibrium \(a_1^* = a_2^* = 1\). C: **Incorrect:** \(1.25\) is not a fixed point of \(a^* = (1 + a^*)/2\). D: **Incorrect:** At \(a_j = 0\), the best response is \(a_i = 0.5 \ne 0\).

**Step 1: Substitute the Nash equilibrium actions \(a_1^* = 1, a_2^* = 1\) into the payoff function \(v_i(a_1, a_2) = a_i - 2a_j + a_i a_j - a_i^2\).** \[v_i(1, 1) = 1 - 2(1) + (1)(1) - (1)^2 = 1 - 2 + 1 - 1 = -1\] **Answer: C (-1)** A: **Incorrect:** Does not match \(1 - 2 + 1 - 1 = -1\). B: **Incorrect:** Sign error; the payoff at \((1, 1)\) is \(-1\). C: **Correct:** Substituting \(a_1 = a_2 = 1\) into \(v_i\) gives \(1 - 2 + 1 - 1 = -1\). D: **Incorrect:** \(0\) is the cooperative payoff when \(a_1 = a_2 = 0\), not the Nash equilibrium payoff.

**Step 1: Maximize the joint (symmetric) payoff \(v(a, a)\) over \(a \in [0, 2]\).** When both parties agree on a common campaign level \(a_1 = a_2 = a \in [0, 2]\), each party's payoff is: \[v_i(a, a) = a - 2a + a^2 - a^2 = -a\] Since \(-a\) is strictly decreasing in \(a\) for \(a \ge 0\), it is maximized on \([0, 2]\) at the boundary: \[a_1 = 0\] **Answer: 0**

By symmetry with Q#126, to maximize joint payoffs \(v_1(a, a) = v_2(a, a) = -a\) over \(a \in [0, 2]\), both parties agree not to run any negative ads: \[a_2 = 0\] **Answer: 0**

When both parties stick to the cooperative agreement \(a_1 = 0, a_2 = 0\) in each stage game, each player's stage-game payoff is: \[v_i(0, 0) = 0 - 2(0) + (0)(0) - 0^2 = 0\] **Answer: D (0)** A: **Incorrect:** \(-1.5\) is not the payoff at \((0, 0)\). B: **Incorrect:** \(v_i(0, 0) = 0\), not \(1\). C: **Incorrect:** \(-1\) is the non-cooperative Nash equilibrium payoff at \((1, 1)\). D: **Correct:** At \(a_1 = a_2 = 0\), \(v_i(0, 0) = 0\).

**Step 1: Find the optimal one-shot deviation \(a_i\) when the other party plays the cooperative level \(a_j = 0\).** When \(a_j = 0\), party \(i\)'s stage payoff is: \[v_i(a_i, 0) = a_i - 2(0) + a_i(0) - a_i^2 = a_i - a_i^2\] Maximizing \(a_i - a_i^2\) with respect to \(a_i\): \[\frac{d}{da_i}(a_i - a_i^2) = 1 - 2a_i = 0 \implies a_i = \frac{1}{2}\] At \(a_i = \frac{1}{2}\), the deviating player gets a one-period payoff of \(\frac{1}{2} - \left(\frac{1}{2}\right)^2 = \frac{1}{4}\). **Answer: B (1/2)** A: **Incorrect:** \(a_i = 1\) gives \(1 - 1^2 = 0\), not \(1/4\). B: **Correct:** Against \(a_j = 0\), the best response is \(a_i = 1/2\), which yields the maximum deviation payoff of \(1/4\). C: **Incorrect:** \(a_i = 1/3\) yields \(1/3 - 1/9 = 2/9 < 1/4\). D: **Incorrect:** \(a_i = 1/4\) yields \(1/4 - 1/16 = 3/16 < 1/4\).

**Step 1: Compare the discounted stream of payoffs from cooperating vs. deviating under grim-trigger punishment.** - **Cooperating** (\(a_1 = a_2 = 0\) forever): Payoff is \(0\) in every period, so total discounted payoff is \(0\). - **Deviating** (playing \(a_i = 1/2\) today to get \(1/4\), followed by reversion to the stage-game Nash equilibrium \(a_1 = a_2 = 1\) with payoff \(-1\) forever after): \[V_{\text{dev}} = \frac{1}{4} + \delta(-1) + \delta^2(-1) + \dots = \frac{1}{4} - \frac{\delta}{1 - \delta}\] **Step 2: Find when cooperation CAN and CANNOT be supported.** Cooperation is sustainable if \(V_{\text{coop}} \ge V_{\text{dev}}\): \[0 \ge \frac{1}{4} - \frac{\delta}{1 - \delta} \iff \frac{\delta}{1 - \delta} \ge \frac{1}{4} \iff 4\delta \ge 1 - \delta \iff 5\delta \ge 1 \iff \delta \ge \frac{1}{5}\] Therefore, the agreement **cannot** be supported for discount factors \(\delta \le \frac{1}{5}\) (strictly \(\delta < 1/5\), represented by **Option B: \(\delta \le \frac{1}{5}\)**). **Answer: B (\(\delta \le \frac{1}{5}\))** A: **Incorrect:** \(\delta \ge 1/5\) is the range where the agreement *can* be supported. B: **Correct:** When \(\delta \le 1/5\) (i.e. below the threshold \(1/5\)), the one-shot deviation gain \(1/4\) outweighs the future punishment \(\frac{\delta}{1-\delta}\). C: **Incorrect:** The critical discount factor is \(1/5\), not \(1/3\). D: **Incorrect:** The critical discount factor is \(1/5\), not \(1/3\).

Let us check the pure-strategy Nash equilibria of the given symmetric game matrix: - **At (Slow, Slow)** with payoffs \((6, 6)\): If Player 1 deviates to `Fast`, Player 1 gets \(2 < 6\). By symmetry, Player 2 also loses by deviating (\(2 < 6\)). Thus, **(Slow, Slow) is a Nash Equilibrium**. - **At (Slow, Fast)** with payoffs \((0, 2)\): Player 1 can deviate from `Slow` (payoff \(0\)) to `Fast` (payoff \(1\)), and Player 2 can deviate from `Fast` (payoff \(2\)) to `Slow` (payoff \(6\)). Thus, **(Slow, Fast) is NOT a Nash Equilibrium**. - (Note: **(Fast, Fast)** with payoffs \((1, 1)\) is also a strict Nash equilibrium since deviating to `Slow` gives \(0 < 1\).) Among the given options, **(Slow, Slow) is an NE** is the only true statement. **Answer: A ((Slow, Slow) is an NE)** A: **Correct:** At (Slow, Slow), neither player wants to deviate because \(6 > 2\). B: **Incorrect:** At (Slow, Fast), both players have profitable unilateral deviations. C: **Incorrect:** (Slow, Fast) is not a Nash equilibrium. D: **Incorrect:** (Slow, Slow) is a Nash equilibrium.

**Step 1: Recall the condition for an Evolutionarily Stable Strategy (ESS) in a symmetric game.** Any **strict symmetric Nash equilibrium** \((s^*, s^*)\)—where \(u(s^*, s^*) > u(s, s^*)\) for all \(s \ne s^*\)—is automatically an **ESS**: 1. **Slow**: \(u(\text{Slow}, \text{Slow}) = 6 > 2 = u(\text{Fast}, \text{Slow})\) \(\implies\) **Slow is a strict NE and therefore an ESS**. 2. **Fast**: \(u(\text{Fast}, \text{Fast}) = 1 > 0 = u(\text{Slow}, \text{Fast})\) \(\implies\) **Fast is a strict NE and therefore an ESS**. Thus, **Both Slow & Fast** are ESS in this game. **Answer: C (Both Slow & Fast)** A: **Incorrect:** Incomplete, because `Fast` is also an ESS. B: **Incorrect:** Incomplete, because `Slow` is also an ESS. C: **Correct:** Both `Slow` (\(6 > 2\)) and `Fast` (\(1 > 0\)) are strict symmetric Nash equilibria and therefore both are ESS. D: **Incorrect:** Both strategies are evolutionarily stable.

The red-majority urn contains **4 red balls and 2 blue balls** (a total of \(4 + 2 = 6\) balls). Therefore, the probability of drawing a blue ball from the red-majority urn is: \[P(\text{Blue} \mid \text{Red-majority}) = \frac{2}{6} = \frac{1}{3}\] **Answer: A (1/3)** A: **Correct:** Out of 6 balls in the red-majority urn, 2 are blue, giving \(2/6 = 1/3\). B: **Incorrect:** \(3/4\) does not match the 6-ball urn composition. C: **Incorrect:** \(1/4\) does not match \(2/6\). D: **Incorrect:** \(2/3 = 4/6\) is the probability of drawing a *red* ball from the red-majority urn (or a blue ball from the blue-majority urn).

**Step 1: List the prior probabilities and conditional likelihoods of drawing a Blue ball.** - Priors: \(P(\text{Red-maj}) = 0.5\), \(P(\text{Blue-maj}) = 0.5\) - Likelihoods: \(P(\text{Blue} \mid \text{Red-maj}) = \frac{2}{6} = \frac{1}{3}\), and \(P(\text{Blue} \mid \text{Blue-maj}) = \frac{4}{6} = \frac{2}{3}\) **Step 2: Apply Bayes' Rule.** \[P(\text{Red-maj} \mid \text{Blue}) = \frac{0.5 \times \frac{1}{3}}{0.5 \times \frac{1}{3} + 0.5 \times \frac{2}{3}} = \frac{1/3}{1/3 + 2/3} = \frac{1}{3}\] **Answer: D (1/3)** A: **Incorrect:** Does not match the 4:2 ball ratio. B: **Incorrect:** Does not match the 4:2 ball ratio. C: **Incorrect:** \(2/3\) is the posterior probability that the urn is *blue-majority* given a blue draw. D: **Correct:** By Bayes' theorem, \(P(\text{Red-maj} \mid \text{Blue}) = \frac{1/3}{1/3 + 2/3} = \frac{1}{3}\).

**Step 1: Identify the single-draw probabilities in the blue-majority urn (with replacement).** In the blue-majority urn (4 Blue, 2 Red): - \(P(\text{Blue} \mid \text{Blue-maj}) = \frac{4}{6} = \frac{2}{3}\) - \(P(\text{Red} \mid \text{Blue-maj}) = \frac{2}{6} = \frac{1}{3}\) **Step 2: Multiply the independent probabilities for the sequence (Blue, Blue, Red).** \[P(B, B, R \mid \text{Blue-maj}) = \frac{2}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{4}{27}\] **Answer: B (4/27)** A: **Incorrect:** \(2 \times 2 \times 1 = 4\), not \(5\). B: **Correct:** \((2/3) \times (2/3) \times (1/3) = 4/27\). C: **Incorrect:** Uses denominator \(5^3 = 125\) instead of \(3^3 = 27\). D: **Incorrect:** Uses denominator \(125\) instead of \(27\).

**Step 1: Identify the single-draw probabilities in the red-majority urn (with replacement).** In the red-majority urn (4 Red, 2 Blue): - \(P(\text{Blue} \mid \text{Red-maj}) = \frac{2}{6} = \frac{1}{3}\) - \(P(\text{Red} \mid \text{Red-maj}) = \frac{4}{6} = \frac{2}{3}\) **Step 2: Multiply the independent probabilities for the sequence (Blue, Blue, Red).** \[P(B, B, R \mid \text{Red-maj}) = \frac{1}{3} \times \frac{1}{3} \times \frac{2}{3} = \frac{2}{27}\] **Answer: A (2/27)** A: **Correct:** \((1/3) \times (1/3) \times (2/3) = 2/27\). B: **Incorrect:** \(5/27\) is not the product of the three draw probabilities. C: **Incorrect:** Uses \(6^3 = 216\) with an incorrect numerator. D: **Incorrect:** Uses \(6^3 = 216\) with an incorrect numerator.

**Step 1: Apply the Law of Total Probability using the results from Q#135 and Q#136.** Since \(P(\text{Blue-maj}) = P(\text{Red-maj}) = \frac{1}{2}\): \[P(B, B, R) = \frac{1}{2} P(B, B, R \mid \text{Blue-maj}) + \frac{1}{2} P(B, B, R \mid \text{Red-maj})\] \[P(B, B, R) = \frac{1}{2}\left(\frac{4}{27}\right) + \frac{1}{2}\left(\frac{2}{27}\right) = \frac{6}{54} = \frac{1}{9}\] **Answer: C (1/9)** A: **Incorrect:** Uses a 5-ball urn denominator. B: **Incorrect:** Uses a 5-ball urn denominator. C: **Correct:** Averaging \(4/27\) and \(2/27\) gives \(3/27 = 1/9\). D: **Incorrect:** Forgets to multiply by the prior probability \(1/2\) (\(4/27 + 2/27 = 6/27 = 2/9\)).

**Step 1: Apply Bayes' Theorem using Q#136 and Q#137.** \[P(\text{Red-maj} \mid B, B, R) = \frac{P(\text{Red-maj})\,P(B, B, R \mid \text{Red-maj})}{P(B, B, R)} = \frac{\frac{1}{2} \times \frac{2}{27}}{\frac{1}{9}} = \frac{1/27}{3/27} = \frac{1}{3}\] **Answer: D (1/3)** A: **Incorrect:** \(2/3\) is the posterior probability that the urn is *blue-majority*. B: **Incorrect:** Does not match Bayes' rule. C: **Incorrect:** Does not match Bayes' rule. D: **Correct:** One Blue and one Red cancel out, leaving a net of one Blue draw, so \(P(\text{Red-maj} \mid B, B, R) = 1/3\).

**Step 1: Read the preference rankings from the two tables (where `1` = most preferred).** - **Men's preferences over Women \((A, B, C, D, E)\)** (right table): - Man **a**: \(B (1) \succ C (2) \succ D (3) \succ E (4) \succ A (5)\) - Man **b**: \(B (1) \succ C (2) \succ E (3) \succ A (4) \succ D (5)\) - Man **c**: \(C (1) \succ E (2) \succ D (3) \succ B (4) \succ A (5)\) - **Women's preferences over Men \((a, b, c)\)** (left table): - Woman **A**: \(a (1) \succ c (2) \succ b (3)\) - Woman **B**: \(c (1) \succ a (2) \succ b (3)\) - Woman **C**: \(a (1) \succ b (2) \succ c (3)\) - Woman **D**: \(b (1) \succ a (2) \succ c (3)\) - Woman **E**: \(a (1) \succ b (2) \succ c (3)\) **Step 2: Run the Men-Proposing Gale–Shapley Algorithm.** - **Round 1**: - Man **a** proposes to his 1st choice, **B**. - Man **b** proposes to his 1st choice, **B**. - Man **c** proposes to his 1st choice, **C**. - Woman **B** compares **a** (rank 2) and **b** (rank 3) \(\implies\) **B holds a** and rejects **b**. - Woman **C** holds **c**. - **Round 2**: - Rejected Man **b** proposes to his 2nd choice, **C**. - Woman **C** compares **b** (rank 2) and **c** (rank 3) \(\implies\) **C holds b** and rejects **c**. - **Round 3**: - Rejected Man **c** proposes to his 2nd choice, **E**. - Woman **E** holds **c**. All three men are now engaged: **(a, B), (b, C), (c, E)**, while **Ms. A** and **Ms. D** remain unmatched (**None**). **Answer: D (None)** A: **Incorrect:** Man `a` is paired with Ms. B. B: **Incorrect:** Man `b` is paired with Ms. C. C: **Incorrect:** Man `c` is paired with Ms. E. D: **Correct:** In the stable matching `(a, B), (b, C), (c, E)`, Ms. A receives no proposals and is paired with None.

From the Gale–Shapley algorithm trace in Q#139: - In Round 1, both Man **a** and Man **b** propose to **Ms. B** (their 1st choice). - **Ms. B** prefers **a** (rank 2) over **b** (rank 3), so she accepts **a** and never receives a proposal from **c** (who gets his 2nd choice **E**). Thus, **Ms. B is paired with a**. **Answer: A (a)** A: **Correct:** Ms. B is paired with Man `a`. B: **Incorrect:** Ms. B rejects Man `b` in favor of Man `a`. C: **Incorrect:** Man `c` is paired with Ms. E. D: **Incorrect:** Ms. B is matched with `a`.

From the Gale–Shapley algorithm trace in Q#139: - In Round 1, **Ms. C** holds Man **c**. - In Round 2, Man **b** (after being rejected by Ms. B) proposes to his 2nd choice, **Ms. C**. - Since **Ms. C** prefers **b** (rank 2) over **c** (rank 3), she accepts **b** and rejects **c**. Thus, **Ms. C is paired with b**. **Answer: B (b)** A: **Incorrect:** Man `a` is paired with Ms. B. B: **Correct:** Ms. C is paired with Man `b`. C: **Incorrect:** Ms. C rejects Man `c` in Round 2 when Man `b` proposes. D: **Incorrect:** Ms. C is matched with `b`.

From the Gale–Shapley algorithm trace in Q#139, the final stable matching is **(a, B), (b, C), (c, E)**. Because there are 5 women and only 3 men, **Ms. D** (along with Ms. A) remains unmatched (**None**). **Answer: D (None)** A: **Incorrect:** Man `a` is paired with Ms. B. B: **Incorrect:** Man `b` is paired with Ms. C. C: **Incorrect:** Man `c` is paired with Ms. E. D: **Correct:** Ms. D is not paired with any man (`None`).

From the Gale–Shapley algorithm trace in Q#139: - After Man **c** is rejected by Ms. C in Round 2, he proposes in Round 3 to his 2nd choice, **Ms. E**, who accepts him. Thus, **Ms. E is paired with c**. **Answer: C (c)** A: **Incorrect:** Man `a` is paired with Ms. B. B: **Incorrect:** Man `b` is paired with Ms. C. C: **Correct:** Ms. E is paired with Man `c`. D: **Incorrect:** Ms. E is matched with `c`.

**Step 1: Recall the addition rule of probability applied to the events \(E = A \cup B\) and \(C\).** \[P((A \cup B) \cup C) = P(A \cup B) + P(C) - P((A \cup B) \cap C)\] **Step 2: Substitute \(P(A \cup B \cup C) = 0.8\), \(P(A \cup B) = 0.6\), and \(P(C) = 0.3\).** \[0.8 = 0.6 + 0.3 - P((A \cup B) \cap C) \implies P((A \cup B) \cap C) = 0.9 - 0.8 = 0.1\] **Answer: B (0.1)** A: **Incorrect:** \(0.2 = 0.8 - 0.6\) is \(P(C \setminus (A \cup B))\), not the intersection. B: **Correct:** \(P((A \cup B) \cap C) = P(A \cup B) + P(C) - P(A \cup B \cup C) = 0.6 + 0.3 - 0.8 = 0.1\). C: **Incorrect:** Exceeds \(P(C) = 0.3\). D: **Incorrect:** \(0.3\) is \(P(C)\), which would only equal the intersection if \(C \subseteq A \cup B\).

**Step 1: Find the percentage of students who like painting but not reading.** Since every student likes at least one of the two activities, \(P(R \cup P) = 100\%\). The percentage of students who like **painting but not reading** is: \[P(P \setminus R) = P(R \cup P) - P(R) = 100\% - 65\% = 35\%\] **Step 2: Multiply by the total number of students (800).** \[\text{Number of students} = 35\% \times 800 = 0.35 \times 800 = 280\] **Answer: C (280)** A: **Incorrect:** \(240\) is \(30\%\) of 800. B: **Incorrect:** \(400 = 50\% \times 800\) is the total number of students who like painting (including those who also like reading). C: **Correct:** \(800 - 0.65(800) = 800 - 520 = 280\) students like painting only. D: **Incorrect:** \(120 = 15\% \times 800\) is the number of students who like *both* reading and painting.

**Step 1: Determine the possible lengths of numbers greater than 10,000 formed from the 6 digits \(\{0, 2, 5, 4, 6, 8\}\) without repetition.** Any 5-digit or 6-digit number (whose leading digit is not `0`) is strictly greater than 10,000: 1. **5-digit numbers**: - First (ten-thousands) digit: can be any of the 5 non-zero digits \(\{2, 5, 4, 6, 8\}\) \(\implies 5\) choices. - Remaining 4 positions: chosen from the remaining 5 digits without repetition \(\implies 5 \times 4 \times 3 \times 2 = 120\) ways. - Total 5-digit numbers \(= 5 \times 120 = 600\). 2. **6-digit numbers**: - First digit: can be any of the 5 non-zero digits \(\implies 5\) choices. - Remaining 5 positions: arranged from the remaining 5 digits in \(5! = 120\) ways. - Total 6-digit numbers \(= 5 \times 120 = 600\). **Step 2: Sum the 5-digit and 6-digit numbers.** \[\text{Total numbers} = 600 + 600 = 1200\] **Answer: D (1200)** A: **Incorrect:** \(600\) counts only the 5-digit numbers and forgets the 6-digit numbers. B: **Incorrect:** \(120 = 5!\) is only the number of 5-digit numbers without `0`. C: **Incorrect:** Undercounts both 5-digit and 6-digit permutations. D: **Correct:** There are \(600\) five-digit numbers and \(600\) six-digit numbers, totaling \(1200\).

**Step 1: Count the total number of ways to choose which 4 students secure the top 4 ranks (in any order).** From 8 students, the number of unordered subsets of 4 students that can occupy the top 4 ranks is: \[\binom{8}{4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70\] (Equivalently, using ordered rankings: \(4! / P(8, 4) = 24 / 1680 = 1 / 70\).) **Step 2: Compute the probability that the top 4 are the 4 prime-numbered students \(\{2, 3, 5, 7\}\).** There is only **1** subset consisting exclusively of the 4 prime-numbered students \(\{2, 3, 5, 7\}\): \[P = \frac{1}{70} \approx 0.014286 \approx 0.014\] **Answer: 0.014**

**Step 1: Calculate the total number of pairs among all 15 offices.** \[\binom{15}{2} = \frac{15 \times 14}{2} = 105\] **Step 2: Subtract the pairs among the 3 smallest offices that do not communicate directly with one another.** \[\binom{3}{2} = \frac{3 \times 2}{2} = 3\] \[\text{Total channels needed} = 105 - 3 = 102\] **Answer: 102**

**Step 1: Treat the 4 students who must sit together as a single combined unit.** We now have **4 teachers + 1 student-block = 5 units** to arrange around a circular table. The number of circular permutations of 5 units is: \[(5 - 1)! = 4! = 24\] **Step 2: Multiply by the internal arrangements of the 4 students within their block.** The 4 students can be arranged among themselves in \(4! = 24\) ways. **Step 3: Compute the total number of seating arrangements.** \[\text{Total ways} = 4! \times 4! = 24 \times 24 = 576\] **Answer: 576**

**Step 1: Compute the total number of ways to select 3 balls from the \(5 + 4 = 9\) balls.** \[\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84\] **Step 2: Subtract the selections that contain NO green balls (i.e., all 3 balls are red).** \[\binom{5}{3} = \frac{5 \times 4}{2 \times 1} = 10\] \[\text{Ways with at least 1 green ball} = 84 - 10 = 74\] **Answer: 74**

**Step 1: Count the choices for each digit position of the 5-digit number using \(\{0, 2, 3, 4, 7\}\) with repetition.** - **1st digit (ten-thousands place)**: Cannot be `0`, so it can be chosen from \(\{2, 3, 4, 7\}\) \(\implies 4\) choices. - **2nd, 3rd, and 4th digits**: Can be any of the 5 digits \(\{0, 2, 3, 4, 7\}\) \(\implies 5 \times 5 \times 5 = 125\) choices. - **5th digit (units place)**: For the number to be divisible by 2 (even), the last digit must be chosen from the even digits \(\{0, 2, 4\}\) \(\implies 3\) choices. **Step 2: Multiply the choices across all 5 positions.** \[\text{Total 5-digit even numbers} = 4 \times 5 \times 5 \times 5 \times 3 = 1500\] **Answer: 1500**

**Step 1: Identify the composition of Box C and Box D.** - **Box C**: 3 white, 2 black (total = 5 balls). - \(P(\text{transfer White}) = \frac{3}{5}\) - \(P(\text{transfer Black}) = \frac{2}{5}\) - **Box D**: Initially 4 white, 3 black (total = 7 balls; after receiving 1 ball, total = 8 balls). **Step 2: Apply the Law of Total Probability to find the probability of drawing a black ball from Box D.** - If a **white** ball is transferred (prob \(\frac{3}{5}\)), Box D has 5 white and 3 black balls \(\implies P(\text{Black} \mid W) = \frac{3}{8}\). - If a **black** ball is transferred (prob \(\frac{2}{5}\)), Box D has 4 white and 4 black balls \(\implies P(\text{Black} \mid B) = \frac{4}{8}\). \[P(\text{Black from D}) = \frac{3}{5} \times \frac{3}{8} + \frac{2}{5} \times \frac{4}{8} = \frac{9 + 8}{40} = \frac{17}{40} = 0.425\] **Answer: 0.425**

**Step 1: Separate the vowels and consonants in the 7-letter word `'ARTICLE'`.** - **Vowels**: `A, I, E` (3 distinct vowels) - **Consonants**: `R, T, C, L` (4 distinct consonants) **Step 2: Group the 3 vowels into a single block.** We now have **4 consonants + 1 vowel-block = 5 units**, which can be arranged in \(5! = 120\) ways. Within the vowel-block, the 3 vowels `A, I, E` can be arranged in \(3! = 6\) ways. **Step 3: Multiply to find the total arrangements.** \[\text{Total ways} = 5! \times 3! = 120 \times 6 = 720\] **Answer: 720**

**Step 1: Find the total number of people \(n\).** If a 3-person committee must include both A and B, the remaining 1 member can be chosen from the remaining \(n - 2\) people in \(\binom{n - 2}{1} = n - 2\) ways. We are given that this number is 7: \[n - 2 = 7 \implies n = 9\] **Step 2: Compute the total number of ways to form a 3-person committee from \(n = 9\) people.** \[\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84\] **Answer: 84**

The bag initially contains **12 gold** and **9 silver** coins (a total of \(12 + 9 = 21\) coins). In the first draw of 3 coins without replacement: - Number of ways to choose 2 gold coins from 12 is \({}^{12}C_2\). - Number of ways to choose 1 silver coin from 9 is \({}^{9}C_1\). - Total ways to choose 3 coins from 21 is \({}^{21}C_3\). Therefore, the probability is: \[\frac{{}^{12}C_2 \times {}^{9}C_1}{{}^{21}C_3}\] **Answer: C (\(\frac{{}^{12}C_2 \times {}^9C_1}{{}^{21}C_3}\))** A: **Incorrect:** Adds \({}^{12}C_2 + {}^9C_1\) instead of multiplying them. B: **Incorrect:** \({}^{12}C_1 \times {}^9C_2\) corresponds to 1 gold and 2 silver coins. C: **Correct:** By the multiplication principle, the favorable outcomes are \({}^{12}C_2 \times {}^9C_1\) divided by \({}^{21}C_3\). D: **Incorrect:** \({}^{12}C_3\) corresponds to drawing 3 gold coins and 0 silver coins.

**Step 1: Determine the coins remaining in the bag after the first draw.** Initially there were 12 gold and 9 silver coins. After removing 2 gold coins and 1 silver coin in the first draw (without replacement), the bag has: - **Gold coins remaining**: \(12 - 2 = 10\) - **Silver coins remaining**: \(9 - 1 = 8\) - **Total coins remaining**: \(10 + 8 = 18\) **Step 2: Find the probability of drawing 4 coins containing "exactly 2 silver coins and at least 1 gold coin".** Since 4 coins are drawn in total and **exactly 2 are silver**, the remaining \(4 - 2 = 2\) coins must be **gold** (which automatically satisfies "at least 1 gold coin"): - Ways to choose 2 silver coins from 8 remaining: \({}^8C_2\) - Ways to choose 2 gold coins from 10 remaining: \({}^{10}C_2\) - Total ways to choose 4 coins from 18 remaining: \({}^{18}C_4\) \[P = \frac{{}^8C_2 \times {}^{10}C_2}{{}^{18}C_4}\] **Answer: B (\(\frac{{}^8C_2 \times {}^{10}C_2}{{}^{18}C_4}\))** A: **Incorrect:** Uses \({}^{10}C_1\) (only 3 coins drawn) and denominator \({}^{21}C_4\) (ignoring that 3 coins were removed). B: **Correct:** With 10 gold and 8 silver coins remaining (18 total), choosing 2 silver and 2 gold gives \(\frac{{}^8C_2 \times {}^{10}C_2}{{}^{18}C_4}\). C: **Incorrect:** Uses the initial counts 9 and 12 in the numerator instead of the remaining counts 8 and 10. D: **Incorrect:** Uses the initial counts 12 and 9 in the numerator.

Let us analyze the two events when selecting a 3-member student panel from **3 Mathematics students and 3 Statistics students**: - **Event \(E_1\)**: The selected panel includes *at least one Mathematics student*. - **Event \(E_2\)**: The selected panel includes *only Statistics students* (i.e., 0 Mathematics students and 3 Statistics students). 1. **Mutually exclusive (Option A is correct)**: Since \(E_1\) requires \(\ge 1\) Mathematics student and \(E_2\) requires \(0\) Mathematics students, \(E_1\) and \(E_2\) cannot occur simultaneously (\(E_1 \cap E_2 = \emptyset\)). Thus, they are **mutually exclusive** (in fact, they are complementary events!). 2. **Independent (Option B is incorrect)**: Because \(P(E_1) > 0\) and \(P(E_2) > 0\) while \(P(E_1 \cap E_2) = 0 \ne P(E_1)P(E_2)\), they are **not independent**. 3. **Favorable outcomes for \(E_2\) (Option C is correct)**: Choosing all 3 Statistics students from the 3 available Statistics students can be done in \(\binom{3}{3} = 1\) way. Thus, there is **only 1 favorable outcome for event \(E_2\)**. **Answer: A, C** A: **Correct:** Having at least one Mathematics student (\(E_1\)) and having only Statistics students (\(E_2\)) are disjoint (complementary) events. B: **Incorrect:** Disjoint events with non-zero probabilities are never independent because \(P(E_1 \cap E_2) = 0 \ne P(E_1)P(E_2)\). C: **Correct:** There are only 3 Statistics students in total, so choosing 3 Statistics students has \(\binom{3}{3} = 1\) outcome. D: **Incorrect:** Option B is false.