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-# COMP182 Homework 9
+# COMP182 Homework 9: Discrete Probability
 
 ## Student Responses
 
-<!-- Write your solutions below. Add sections for each problem as needed. -->
-<!-- Use proper mathematical notation where applicable. -->
+---
 
-[Your solutions here]
+## 1. Events and Their Probabilities
 
+### Problem 1 [24 pts]
+
+We randomly select a permutation of the 26 lowercase letters of the English alphabet. The total number of permutations is 26!.
+
+**(a)** The first 13 letters of the permutation are in alphabetical order.
+
+We need to choose which 13 letters occupy the first 13 positions (they must be arranged in alphabetical order, so there is only 1 way to arrange them once chosen), and the remaining 13 letters can be in any order in the last 13 positions.
+
+- Number of ways to choose which 13 letters go in the first 13 positions: C(26, 13)
+- Arrangement of those 13 letters in the first 13 positions: 1 (must be alphabetical)
+- Arrangement of the remaining 13 letters in the last 13 positions: 13!
+- Favorable outcomes: C(26, 13) × 13!
+
+P = C(26, 13) × 13! / 26!
+= [26! / (13! × 13!)] × 13! / 26!
+= 1 / 13!
+
+**P = 1/13!**
+
+**(b)** *a* is the first letter and *z* is the last letter.
+
+Fix *a* in position 1 and *z* in position 26. The remaining 24 letters can be arranged in any order in the remaining 24 positions.
+
+P = 24! / 26! = 1 / (26 × 25)
+
+**P = 1/650**
+
+**(c)** *a* and *z* are next to each other in the permutation.
+
+Treat {a, z} as a single block. Then we have 25 objects to arrange (the block plus 24 other letters). These 25 objects can be arranged in 25! ways. Within the block, a and z can appear in 2 orders (az or za).
+
+Favorable outcomes: 2 × 25!
+
+P = 2 × 25! / 26! = 2/26
+
+**P = 1/13**
+
+**(d)** *a* and *b* are not next to each other in the permutation.
+
+Using the complement:
+
+P(a and b not adjacent) = 1 − P(a and b adjacent)
+
+By the same reasoning as part (c), P(a and b adjacent) = 2 × 25! / 26! = 1/13.
+
+P = 1 − 1/13
+
+**P = 12/13**
+
+**(e)** *a* and *z* are separated by at least 23 letters in the permutation.
+
+If *a* is at position i and *z* is at position j, the number of letters between them is |i − j| − 1. We need |i − j| − 1 ≥ 23, i.e., |i − j| ≥ 24.
+
+Since positions range from 1 to 26, the possible cases are:
+
+- |i − j| = 24: ordered pairs (i, j) are (1,25), (25,1), (2,26), (26,2) → 4 pairs
+- |i − j| = 25: ordered pairs (i, j) are (1,26), (26,1) → 2 pairs
+
+Total favorable ordered pairs for (position of a, position of z): 6
+
+Total ways to assign positions to a and z: 26 × 25 = 650
+
+P = 6/650
+
+**P = 3/325**
+
+**(f)** *z* precedes both *a* and *b* in the permutation.
+
+Consider only the relative ordering of the three letters z, a, and b. All other letters are irrelevant. There are 3! = 6 equally likely orderings of these three letters. We want z to come first. The number of orderings where z is first is 2! = 2 (since a and b can be in either order after z).
+
+P = 2/6
+
+**P = 1/3**
+
+---
+
+### Problem 2 [9 pts]
+
+All days of the week are equally likely (7 days).
+
+**(a)** Probability that two people were born on the same day of the week.
+
+The first person can be born on any day. The second person must be born on the same day, with probability 1/7.
+
+**P = 1/7**
+
+**(b)** Probability that in a group of n people, at least two share a birthday (day of week).
+
+Using the complement:
+
+P(at least two share) = 1 − P(all different)
+
+For n > 7: By the Pigeonhole Principle, at least two must share a day, so P = 1.
+
+For n ≤ 7: The number of ways all n people are born on different days is 7 × 6 × 5 × ... × (7 − n + 1) = 7!/(7 − n)!. The total number of outcomes is 7^n.
+
+**P(at least two share a day) = 1 − 7! / ((7 − n)! × 7^n)** for n ≤ 7, and **P = 1** for n > 7.
+
+**(c)** Find the smallest n such that this probability exceeds 1/2.
+
+We compute P(all different) = 7!/((7−n)! × 7^n) for successive values of n:
+
+- n = 1: P(all different) = 7/7 = 1, so P(at least two same) = 0
+- n = 2: P(all different) = (7 × 6)/7^2 = 42/49 = 6/7, so P(at least two same) = 1/7 ≈ 0.143
+- n = 3: P(all different) = (7 × 6 × 5)/7^3 = 210/343, so P(at least two same) = 133/343 ≈ 0.388
+- n = 4: P(all different) = (7 × 6 × 5 × 4)/7^4 = 840/2401, so P(at least two same) = 1561/2401 ≈ 0.650
+
+Since 0.388 < 0.5 < 0.650, the probability first exceeds 1/2 at **n = 4**.
+
+---
+
+### Problem 3 [4 pts]
+
+**Show that if E and F are independent events, then E̅ and F̅ are independent events.**
+
+Since E and F are independent, we have P(E ∩ F) = P(E)P(F).
+
+We need to show P(E̅ ∩ F̅) = P(E̅)P(F̅).
+
+By De Morgan's Law: E̅ ∩ F̅ = (E ∪ F)̅
+
+Therefore:
+
+P(E̅ ∩ F̅) = P((E ∪ F)̅)
+= 1 − P(E ∪ F)
+= 1 − [P(E) + P(F) − P(E ∩ F)]
+= 1 − P(E) − P(F) + P(E ∩ F)
+= 1 − P(E) − P(F) + P(E)P(F)     [using independence of E and F]
+= (1 − P(E))(1 − P(F))
+= P(E̅) × P(F̅)
+
+Since P(E̅ ∩ F̅) = P(E̅)P(F̅), the events E̅ and F̅ are independent. ∎
+
+---
+
+### Problem 4 [6 pts]
+
+Given: P(A) = 2/3, P(B) = 1/2.
+
+**(a)** Bounds on P(A ∩ B):
+
+**Largest:** P(A ∩ B) ≤ min(P(A), P(B)) = min(2/3, 1/2) = **1/2**.
+
+*Example:* Let S = {1, 2, 3, 4, 5, 6} with uniform probability. Let A = {1, 2, 3, 4} and B = {1, 2, 3}. Then P(A) = 4/6 = 2/3, P(B) = 3/6 = 1/2, and A ∩ B = {1, 2, 3}, so P(A ∩ B) = 3/6 = 1/2. This occurs when B ⊆ A.
+
+**Smallest:** By inclusion-exclusion, P(A ∪ B) ≤ 1, so P(A) + P(B) − P(A ∩ B) ≤ 1, giving P(A ∩ B) ≥ P(A) + P(B) − 1 = 2/3 + 1/2 − 1 = **1/6**.
+
+*Example:* Let S = {1, 2, 3, 4, 5, 6} with uniform probability. Let A = {1, 2, 3, 4} and B = {4, 5, 6}. Then P(A) = 4/6 = 2/3, P(B) = 3/6 = 1/2, and A ∩ B = {4}, so P(A ∩ B) = 1/6.
+
+**(b)** Bounds on P(A ∪ B):
+
+Since P(A ∪ B) = P(A) + P(B) − P(A ∩ B):
+
+**Largest:** Minimize P(A ∩ B). Using the minimum value P(A ∩ B) = 1/6:
+P(A ∪ B) = 2/3 + 1/2 − 1/6 = 4/6 + 3/6 − 1/6 = 6/6 = **1**.
+
+*Example:* Same as the smallest intersection example: A = {1, 2, 3, 4}, B = {4, 5, 6}. Then A ∪ B = {1, 2, 3, 4, 5, 6} = S, so P(A ∪ B) = 1.
+
+**Smallest:** Maximize P(A ∩ B). Using the maximum value P(A ∩ B) = 1/2:
+P(A ∪ B) = 2/3 + 1/2 − 1/2 = **2/3**.
+
+*Example:* Same as the largest intersection example: A = {1, 2, 3, 4}, B = {1, 2, 3}. Then A ∪ B = {1, 2, 3, 4}, so P(A ∪ B) = 4/6 = 2/3.
+
+---
+
+### Problem 5 [12 pts]
+
+We perform n independent Bernoulli trials with probability of success p on each trial. The number of successes follows a Binomial(n, p) distribution.
+
+**(a)** Probability of no success (0 successes):
+
+P(X = 0) = C(n, 0) × p^0 × (1 − p)^n
+
+**P = (1 − p)^n**
+
+**(b)** Probability of at least one success:
+
+P(X ≥ 1) = 1 − P(X = 0)
+
+**P = 1 − (1 − p)^n**
+
+**(c)** Probability of at most one success (0 or 1 successes):
+
+P(X ≤ 1) = P(X = 0) + P(X = 1)
+= C(n, 0)p^0(1−p)^n + C(n, 1)p^1(1−p)^(n−1)
+
+**P = (1 − p)^n + np(1 − p)^(n−1)**
+
+**(d)** Probability of at least two successes:
+
+P(X ≥ 2) = 1 − P(X ≤ 1) = 1 − [(1 − p)^n + np(1 − p)^(n−1)]
+
+**P = 1 − (1 − p)^n − np(1 − p)^(n−1)**
+
+---
+
+## 2. Random Variables
+
+### Problem 1 [6 pts]
+
+We roll a fair die until a 6 comes up. Let X be the number of rolls.
+
+**(a)** The probability that we roll the die n times:
+
+For X = n, the first n − 1 rolls must not be 6 (each with probability 5/6), and the n-th roll must be 6 (probability 1/6).
+
+**P(X = n) = (5/6)^(n−1) × (1/6)** for n = 1, 2, 3, ...
+
+**(b)** Expected number of rolls:
+
+This is a geometric distribution with success probability p = 1/6.
+
+E(X) = Σ_{n=1}^∞ n × (5/6)^(n−1) × (1/6)
+
+Using the formula for geometric expectation, E(X) = 1/p:
+
+**E(X) = 1/(1/6) = 6**
+
+*Derivation:* Let q = 5/6. Then:
+E(X) = (1/6) × Σ_{n=1}^∞ n × q^(n−1) = (1/6) × 1/(1−q)^2 = (1/6) × 1/(1/6)^2 = (1/6) × 36 = 6.
+
+---
+
+### Problem 2 [6 pts]
+
+Given:
+- P(S = 1) = 1/3, P(S = 0) = 2/3
+- P(R = 0 | S = 0) = 0.9, P(R = 1 | S = 0) = 0.1
+- P(R = 1 | S = 1) = 0.8, P(R = 0 | S = 1) = 0.2
+
+**(a)** Find P(R = 0).
+
+By the law of total probability:
+
+P(R = 0) = P(R = 0 | S = 0) × P(S = 0) + P(R = 0 | S = 1) × P(S = 1)
+= (9/10)(2/3) + (2/10)(1/3)
+= 18/30 + 2/30
+= 20/30
+
+**P(R = 0) = 2/3**
+
+**(b)** Find P(S = 0 | R = 0).
+
+By Bayes' theorem:
+
+P(S = 0 | R = 0) = P(R = 0 | S = 0) × P(S = 0) / P(R = 0)
+= (9/10)(2/3) / (2/3)
+= (18/30) / (20/30)
+= 18/20
+
+**P(S = 0 | R = 0) = 9/10**
+
+---
+
+### Problem 3 [6 pts]
+
+**Show that E(Z) ≤ E(X) + E(Y), where Z(s) = max(X(s), Y(s)) and X, Y are nonnegative.**
+
+Since X(s) ≥ 0 and Y(s) ≥ 0 for all s ∈ S, we show that max(X(s), Y(s)) ≤ X(s) + Y(s) for all s ∈ S:
+
+- **Case 1:** X(s) ≥ Y(s). Then max(X(s), Y(s)) = X(s) ≤ X(s) + Y(s), since Y(s) ≥ 0.
+- **Case 2:** Y(s) > X(s). Then max(X(s), Y(s)) = Y(s) ≤ X(s) + Y(s), since X(s) ≥ 0.
+
+In both cases, Z(s) ≤ X(s) + Y(s).
+
+Since Z(s) ≤ X(s) + Y(s) for all s ∈ S, by the monotonicity property of expectation:
+
+E(Z) ≤ E(X + Y) = E(X) + E(Y)
+
+where the last equality follows from linearity of expectation. ∎
+
+---
+
+### Problem 4 [3 pts]
+
+Two fair coins are flipped. Let X = number of heads, Y = number of tails.
+
+Sample space: S = {HH, HT, TH, TT}, each outcome with probability 1/4.
+
+| Outcome | X | Y |
+|---------|---|---|
+| HH | 2 | 0 |
+| HT | 1 | 1 |
+| TH | 1 | 1 |
+| TT | 0 | 2 |
+
+Note that X + Y = 2 for all outcomes.
+
+**X and Y are NOT independent.**
+
+*Proof:* We check the independence condition P(X = a, Y = b) = P(X = a) × P(Y = b):
+
+- P(X = 0) = 1/4, P(Y = 0) = 1/4
+- P(X = 0, Y = 0) = P(∅) = 0 (since X = 0 and Y = 0 cannot occur simultaneously — if X = 0 then both coins are tails, so Y = 2, not 0)
+- P(X = 0) × P(Y = 0) = (1/4)(1/4) = 1/16
+
+Since P(X = 0, Y = 0) = 0 ≠ 1/16 = P(X = 0)P(Y = 0), X and Y are **not independent**. ∎
+
+---
+
+### Problem 5 [5 pts]
+
+A fair die is rolled 10 times. Let X = number of times a 6 appears.
+
+X follows a Binomial distribution with n = 10 and p = P(rolling a 6) = 1/6.
+
+The variance of a binomial random variable is:
+
+V(X) = np(1 − p) = 10 × (1/6) × (5/6) = 50/36
+
+**V(X) = 25/18**
+
+---
+
+### Problem 6 [6 pts]
+
+**Example showing V(X + Y) ≠ V(X) + V(Y) when X and Y are not independent.**
+
+Let a fair coin be flipped once. Define:
+- X = 1 if heads, X = 0 if tails
+- Y = X (i.e., Y is the same random variable as X)
+
+X and Y are not independent since knowing X completely determines Y.
+
+**Compute the variances:**
+
+E(X) = 1/2, so V(X) = E(X²) − (E(X))² = 1/2 − 1/4 = 1/4.
+
+Similarly, V(Y) = 1/4.
+
+So V(X) + V(Y) = 1/4 + 1/4 = **1/2**.
+
+**Now compute V(X + Y):**
+
+Since Y = X, we have X + Y = 2X.
+
+V(X + Y) = V(2X) = 4 × V(X) = 4 × 1/4 = **1**.
+
+Since V(X + Y) = 1 ≠ 1/2 = V(X) + V(Y), this demonstrates that the variance of the sum is not necessarily equal to the sum of the variances when the random variables are not independent.
+
+*Note:* The general formula is V(X + Y) = V(X) + V(Y) + 2Cov(X, Y). Here Cov(X, Y) = Cov(X, X) = V(X) = 1/4 ≠ 0, which is why the equality fails.
+
+---
+
+### Problem 7 [4 pts]
+
+A biased coin with P(heads) = 0.6 is tossed n times. Let X = number of tails.
+
+- P(tails) = 1 − 0.6 = 0.4
+- E(X) = μ = n × 0.4 = 2n/5
+- V(X) = σ² = n × 0.4 × 0.6 = 0.24n = 6n/25
+
+We want an upper bound on P(|X − μ| ≥ √n).
+
+By Chebyshev's inequality: P(|X − μ| ≥ k) ≤ σ²/k²
+
+Setting k = √n:
+
+P(|X − 2n/5| ≥ √n) ≤ V(X)/n = (6n/25)/n = 6/25
+
+**P(|X − μ| ≥ √n) ≤ 6/25 = 0.24**
+
+---
+
+### Problem 8 [9 pts]
+
+The probability that x is the i-th element in a list of n distinct integers is p_i = i/(n(n+1)) for i = 1, 2, ..., n.
+
+**First, verify the total probability:**
+
+Σ_{i=1}^n p_i = Σ_{i=1}^n i/(n(n+1)) = [n(n+1)/2] / (n(n+1)) = 1/2
+
+So the probability that x is in the list is 1/2, and the probability that x is not in the list is 1 − 1/2 = 1/2.
+
+**Linear search behavior:**
+- If x is at position i, the algorithm makes i comparisons to find it.
+- If x is not in the list, the algorithm makes n comparisons (checking every element).
+
+**Expected number of comparisons:**
+
+E = Σ_{i=1}^n i × p_i + n × P(x not in list)
+
+= Σ_{i=1}^n i × [i/(n(n+1))] + n × (1/2)
+
+= [1/(n(n+1))] × Σ_{i=1}^n i² + n/2
+
+Using the formula Σ_{i=1}^n i² = n(n+1)(2n+1)/6:
+
+= [1/(n(n+1))] × [n(n+1)(2n+1)/6] + n/2
+
+= (2n+1)/6 + n/2
+
+= (2n+1)/6 + 3n/6
+
+= (2n + 1 + 3n)/6
+
+**E = (5n + 1)/6**