fixed solutions listed in internal (12/24/2024)

Author: Nguyen-HanhNong

src/content/questions/comp2804/2014-fall-final/23/solution.md

@@ -1,11 +1,7 @@
-So um, this algorithm pretty much counts the number of attempts we make until we get $HHH$ or $TTT$
+Since we have an endless algorithm (i.e the algorithm will run forever until it finds a solution), we can model this problem using a geometric distribution.
 
-We have a $ \frac{1}{8} $ chance of flipping an $HHH$
+By the geometric distribution, the expected value of the number of attempts until we get a success is $\frac{1}{p}$, where $p$ is the probability of terminating the algorithm at each step.
 
-We have a $ \frac{1}{8} $ chance of flipping an $TTT$
+In this case, the probability of terminating the algorithm is either getting three heads in a row or getting three tails in a row. The probability of getting three heads in a row is $\frac{1}{2^3} = \frac{1}{8}$, and the probability of getting three tails in a row is also $\frac{1}{8}$. Therefore, the probability of terminating the algorithm at each step is $\frac{1}{8} + \frac{1}{8} = \frac{1}{4}$.
 
-This gives us a $ \frac{1}{4} $ chance of flipping either $HHH$ or $TTT$
-
-By the geometric distribution, the expected value of the number of attempts until we get a success is $ \frac{1}{p} $
-
-Therefore, the expected value of $X$ is $ \frac{1}{ \frac{1}{4}} = 4 $
+Now, we can calculate the expected value using the formula for expected value for geometric distribution: $E(X) = \frac{1}{p} = \frac{1}{\frac{1}{4}} = 4$.

src/content/questions/comp2804/2015-fall-final/7/solution.md

@@ -1,9 +1,13 @@
-Let's write out the possibilities I guess
+# Recursive Breakdown
 
-<ul>
-    <li> $ b, S_{n-1} $
-    <li> $ c, S_{n-1} $
-    <li> $ a, 3^{n-1} $ because we could have our first $a$ here, and the rest of the string can be any of the 3 characters
-</ul>
+Consider two cases based on the **first character** of the string:
 
-$ S_n = 2 \cdot S\_{n-1} + 3^{n-1} $
+1. **If the first character is 'a':**
+
+The remaining $n - 1$ characters can be any string of length $n - 1$ using **a, b, c**. The number of such strings is $3^{n-1}$.
+
+2. **If the first character is 'b' or 'c':**
+
+The string must still contain at least one 'a' in the remaining $n - 1$ characters. This is equivalent to counting valid strings of length $n - 1$ containing at least one 'a', which is $S_{n-1}$. Since the first character can be either **'b'** or **'c'** (2 choices), the contribution is: $2 \cdot S_{n-1}$
+
+Adding both contributions: $S_n = 2 \cdot S_{n-1} + 3^{n-1}$

src/content/questions/comp2804/2017-fall-final/20/solution.md

@@ -1,22 +1,43 @@
-Let's explain why some are wrong and one is right
-
-<ul>
-    <li> ${\sum_{k=1}^{100}} k \cdot \frac{k^{2}}{100^{2}}$ <br/> 
-    This equation says, \enquote{The probability that k is the max value and each value has 1 to k values to be} <br/> 
-    This doesn't work because it's saying that the max value is k if $a$ is 1 and $b$ is 1
-    <li> ${\sum_{k=1}^{100}} k \cdot \frac{k(k-1)}{100^{2}}$ <br/> 
-    This equation says, \enquote{a can be any value from 1 to k and b can be any value from 1 to k-1} <br/> 
-    This doesn't work because it's saying that the max value is k if $a$ is $k-1$ and $b$ is $k-2$
-    <li> ${\sum_{k=1}^{100}} k \cdot \frac{2k}{100^{2}}$ <br/> 
-    This equation says, \enquote{a can be any value from 1 to k and b can be any value from 1 to k} <br/> 
-    This doesn't work because it's saying that the max value is k if $a$ is 1 and $b$ is 1
-    <li> ${\sum_{k=1}^{100}} k \cdot bigg( \frac{1+2(k-1)}{100^{2}}bigg)$ <br/> 
-    This one actually has multiple possibilities <br/> 
-    <ul>
-        <li> $a = k$ and $b = k$: 1 possibility
-        <li> $a = k$ and $b$ is any value less than k: $k(k-1)$ possibilities
-        <li> $a$ is any value less than k and $b = k$: $(k-1)k$ possibilities
-    </ul>
-    Summing them up, we get $1 + 2 \cdot k(k-1) $ 
-    Each of the tree posibilities above makes sure that the max value is k
-</ul>
+## Goal of Problem
+
+We want to calculate the **expected value** of the random variable: $X = \max(a, b)$ where:
+
+- $ a $ and $ b $ are chosen independently and uniformly from the set $ \{1, 2, \ldots, 100\} $.
+
+---
+
+## **Key Observation**
+
+The maximum of two independent uniformly random variables, $ a $ and $ b $, takes the value $ k $ if:
+
+- At least one of the two numbers is equal to $ k $.
+- The other number is **less than or equal to $ k $**.
+
+### **Probability that $ X = k $**
+
+For a maximum of $ k $ to occur:
+
+1. At least one of $ a $ or $ b $ must be equal to $ k $.
+2. The other value must be $ \leq k $.
+
+The total number of outcomes for $ (a, b) $ is: $100^2$ since $ a $ and $ b $ are chosen from $ \{1, 2, \ldots, 100\}$.
+
+The favorable outcomes for $ X = k $ are:
+
+- $ a = k $ and $ b \leq k $: $ k $ outcomes.
+- $ b = k $ and $ a \leq k $: $ k $ outcomes.
+- $ a = b = k $ is counted twice, so subtract 1.
+
+Total favorable outcomes: $2k - 1$
+
+This means the probability that $ X = k $ is: $P(X = k) = \frac{|X = k|}{|S|} = \frac{2k - 1}{100^2}$.
+
+---
+
+## **Substituting Expected Value Formula**
+
+By the definition of expected value: $E(X) = \sum\_{k=1}^{100} k \cdot P(X = k)$
+
+Substituting the probability: $E(X) = \sum\_{k=1}^{100} k \cdot \frac{2k - 1}{100^2}$
+
+We can rewrite $2k-1$ as $1 + 2(k-1)$. They result in the same sum, so we get: $E(X) = \sum\_{k=1}^{100} k \cdot \left(\frac{1 + 2(k-1)}{100^2}\right)$

src/content/questions/comp2804/2017-fall-final/21/solution.md

@@ -1,25 +1,65 @@
-Before drawing a tree, let's think for a second
+## **Counterexample to Show $ X $ and $ Y $ Are Not Independent**
 
-We just need to prove whether the following is true for some value $x$ and $y$: $ Pr(X=i \cap Y=j) = Pr(X=i) \cdot Pr(Y=j) $
+We want to demonstrate mathematically that the random variables:
 
-Let's do it for X=2 and Y=2
+- $ X = i + j $ (sum of two six-sided dice rolls)
+- $ Y = i - j $ (difference between two six-sided dice rolls)
 
-$ |S| = 36 $
+are **not independent** by showing an example where the independence condition $P(X = x \land Y = y) = P(X = x) \cdot P(Y = y)$ **fails** for certain values of $ x $ and $ y $.
 
-$X=2$ for the pair $ (1,1) $
+---
 
-$ Pr(X=2) = \frac{1}{36} $
+## **Choose Specific Values**
 
-$Y=5$ for the pairs $ (6,1), (1,6) $
+Let:
 
-$ Pr(Y=5) = \frac{2}{36} $
+- $ X = 4 $ (sum of the two dice is 4)
+- $ Y = 2 $ (difference between the two dice is 2)
 
-Now, let's find the case of $ X=2 \cap Y=5 $
+We will compute:
 
-When we look at overlapping pairs in $X=2$ and $Y=5$, we see that there is no intersection
+1. $ P(X = 4 \land Y = 2) $
+2. $ P(X = 4) $
+3. $ P(Y = 2) $
+4. Check if the independence formula holds.
 
-This means that $ Pr(X=2 \cap Y=5) = 0 $
+---
 
-$ Pr(X=2 \cap Y=5) = Pr(X=2) \cdot Pr(Y=5) $
+## **Joint Probability $ P(X = 4 \land Y = 2) $**
 
-Since the equation is false, X and Y are not independent.
+For $ X = 4 $ and $ Y = 2 $:
+
+- The sum $ i + j = 4 $
+- The difference $ i - j = 2 $
+
+Possible pairs $(i, j)$ that satisfy both conditions: $ (3,1) $
+
+There is **1 valid outcome** out of $ 36 $ total outcomes, so $P(X = 4 \land Y = 2) = \frac{1}{36}$
+
+---
+
+## **Marginal Probability $ P(X = 4) $**
+
+Possible pairs where the sum is 4: $ (1,3), (2,2), (3,1) $
+
+There are **3 valid outcomes** out of $36$ total outcomes, so $P(X = 4) = \frac{3}{36} = \frac{1}{12}$
+
+---
+
+## **Marginal Probability $ P(Y = 2) $**
+
+Possible pairs where the difference is 2: $ (3,1), (4,2), (5,3), (6,4) $
+
+There are **4 outcomes** out of $36$ total outcomes, so $P(Y = 2) = \frac{4}{36} = \frac{1}{9}$
+
+## **Check Independence Condition**
+
+If $ X $ and $ Y $ were independent, we would expect: $P(X = 4 \land Y = 2) = P(X = 4) \cdot P(Y = 2)$
+
+Substituting the probabilities: $\frac{1}{36} \stackrel{?}{=} \frac{1}{12} \cdot \frac{1}{9}$
+
+Simplify the right-hand side: $\frac{1}{12} \cdot \frac{1}{9} = \frac{1}{108}$
+
+Comparing the two probabilities: $\frac{1}{36} \neq \frac{1}{108}$
+
+Clearly, the probabilities **do not match**, which means that the random variables $ X $ and $ Y $ are **not independent**.

src/content/questions/comp2804/2018-fall-final/11/solution.md

@@ -1,36 +1,18 @@
-Okay, there are two ways to go about this
+## **Answer by Observation**
 
-<ul>
-    <li> The first is to realize that the chance the coin comes up a certain face an odd or even number of times is straight up $ \frac{1}{2} $ <br/> 
-    Screw drawing out the whole 5-level deep recursive tree. We can just draw out a 3-level deep tree <br/> 
-    <img src="/images/comp2804/2018-fall-final/11/image.png"/> <br/> 
-    If you notice, our first flip gets us a head or a tail, which is a 50/50 chance. <br/> 
-    The first flip doubles the possibilities <br/> 
-    If after flipping a heads we flip a heads again, this creates an even number of tails BUT the probability is kept the same by the fact that another possibility is HT <br/> 
-    If after flipping a tails we flip a tails, it's even BUT the counter is that we could land a TH, which brings the probability back to 50/50 <br/> 
-    Ever time we flip the coin to get an odd number of heads, we could just as easily have gotten an even number of tails <br/> 
-    Some of the exam questions are like this, and it's not really worth the time to brute force calculate <br/> 
-    It's as intuitive as it is unintuitive
-    <li> Second, we can draw out a tree and realize that the probability of getting an odd number of heads is the same as getting an even number of heads
-    <li> Third, let's just waste brainpower to solve this <br/> 
-    <ul>
-        <li> Let S be all possible outcomes of the coin flips: $ |S| = 2^5 = 32 $
-        <li> Let A represent the event that the coin comes up heads once <br/> 
-        We choose 1 position out of the 5 for the head: $ \binom{5}{1} $ <br/> 
-        The remaining 4 positions are for the tails: 1 <br/> 
-        $ |A| = \binom{5}{1} $
-        <li> Let B represent the event that the coin comes up heads 3 times <br/> 
-        We choose 3 positions out of the 5 for the heads: $ \binom{5}{3} $ <br/> 
-        The remaining 2 positions are for the tails: 1 <br/> 
-        $ |B| = \binom{5}{3} $
-        <li> Let C represent the event that the coin comes up heads 5 times <br/> 
-        We choose 5 positions out of the 5 for the heads: $ \binom{5}{5} $ <br/> 
-        The remaining 0 positions are for the tails: 1 <br/> 
-        $ |C| = \binom{5}{5} $
-    </ul>
-    $ Pr(\text{odd number of heads}) = Pr(A) + Pr(B) + Pr(C) $<br/>  
-    $ Pr(\text{odd number of heads}) = \frac{ \binom{5}{1} + \binom{5}{3} + \binom{5}{5} }{32} $ <br/> 
-    $ Pr(\text{odd number of heads}) = \frac{5 + 10 + 1}{32} $ <br/> 
-    $ Pr(\text{odd number of heads}) = \frac{16}{32} $ <br/> 
-    $ Pr(\text{odd number of heads}) = \frac{1}{2} $
-</ul>
+Each flip of a fair coin has two equally likely outcomes:
+
+- $ H $ (Heads)
+- $ T $ (Tails)
+
+When flipping 5 coins:
+
+- The total number of possible outcomes is: $2^5 = 32$
+- Half of these outcomes will have an **odd number of heads**, and the other half will have an **even number of heads**.
+
+### **Why Are They Evenly Split?**
+
+- For every outcome with an **odd number of heads**, there is a corresponding outcome with an **even number of heads** by flipping one coin from heads to tails (or vice versa).
+- This creates a **1-to-1 correspondence** between odd and even outcomes.
+
+Thus, the probability of an **odd number of heads** is: $P(\text{Odd number of heads}) = \frac{1}{2}$

src/content/questions/comp2804/2018-fall-final/13/solution.md

@@ -1,18 +1,111 @@
-<ul>
-    <li> Let's determine $ B $ <br/> 
-    What we care about is when we get an $a$ so let's map that out with a table <br/> 
-    <img src="/images/comp2804/2018-fall-final/13/image.png"/> <br/> 
-    $ |B| = 11 $ <br/> 
-    $ Pr(B) = \frac{11}{36} $
-    <li> Let's determine $ A \cap B $ <br/> 
-    We only care about the cases when both rolls result in the same letter AND at least one of the rolls result in the letter $a$ <br/> 
-    ![alt text](image-1.png) <br/> 
-    $ |A \cap B| = 1 $ <br/> 
-    $ Pr(A \cap B) = \frac{1}{36} $
-</ul>
+In this case, our dice has special properties compared to a normal 6-sided dice. <br />
+We can define the dice as having the following results: {a, b, b, c, c, c} <br />
+Now, all we have to do is determine $Pr(A \cap B)$ and $Pr(B)$ to determine $Pr(A|B)$ using the independence formula. <br />
 
-$ Pr(A|B) = \frac{ Pr(A \cap B) }{ Pr(B) } $
+$Pr(B)$ represents the probability of at least one of the rolls resulting in the letter $a$. To make this easier to visualize, we can draw out a table of all possible outcomes between the two dice rolls. <br />
+$|S|$ = the sample space of all possible rolls = $6 \times 6 = 36$ <br />
 
-$ Pr(A|B) = \frac{ \frac{1}{36} }{ \frac{11}{36} } $
+  <table style="
+    border-collapse: collapse; 
+    margin: 0 auto; 
+    table-layout: fixed; 
+    width: 600px;">
+    <caption style="
+      font-size: 1.25rem; 
+      font-weight: bold; 
+      margin-bottom: 10px;">
+      All Possible Rolls for Two Custom 6-Sided Dice
+    </caption>
+    <thead>
+      <tr>
+        <th style="
+          border: 1px solid #ccc; 
+          padding: 10px; 
+          text-align: center; 
+          background-color: #f2f2f2;"></th>
+        <th style="
+          border: 1px solid #ccc; 
+          padding: 10px; 
+          text-align: center; 
+          background-color: #f2f2f2;">Die 2: a</th>
+        <th style="
+          border: 1px solid #ccc; 
+          padding: 10px; 
+          text-align: center; 
+          background-color: #f2f2f2;">Die 2: b</th>
+        <th style="
+          border: 1px solid #ccc; 
+          padding: 10px; 
+          text-align: center; 
+          background-color: #f2f2f2;">Die 2: b</th>
+        <th style="
+          border: 1px solid #ccc; 
+          padding: 10px; 
+          text-align: center; 
+          background-color: #f2f2f2;">Die 2: c</th>
+        <th style="
+          border: 1px solid #ccc; 
+          padding: 10px; 
+          text-align: center; 
+          background-color: #f2f2f2;">Die 2: c</th>
+        <th style="
+          border: 1px solid #ccc; 
+          padding: 10px; 
+          text-align: center; 
+          background-color: #f2f2f2;">Die 2: c</th>
+      </tr>
+    </thead>
+    <tbody>
+      <tr>
+        <th style="
+          border: 1px solid #ccc; 
+          padding: 10px; 
+          text-align: center; 
+          font-weight: bold; 
+          background-color: #f2f2f2;">Die 1: a</th>
+        <td style="border: 1px solid #ccc; padding: 10px; text-align: center;">(a,a)</td>
+        <td style="border: 1px solid #ccc; padding: 10px; text-align: center;">(a,b)</td>
+        <td style="border: 1px solid #ccc; padding: 10px; text-align: center;">(a,b)</td>
+        <td style="border: 1px solid #ccc; padding: 10px; text-align: center;">(a,c)</td>
+        <td style="border: 1px solid #ccc; padding: 10px; text-align: center;">(a,c)</td>
+        <td style="border: 1px solid #ccc; padding: 10px; text-align: center;">(a,c)</td>
+      </tr>
+      <tr>
+        <th style="
+          border: 1px solid #ccc; 
+          padding: 10px; 
+          text-align: center; 
+          font-weight: bold; 
+          background-color: #f2f2f2;">Die 1: b</th>
+        <td style="border: 1px solid #ccc; padding: 10px; text-align: center;">(b,a)</td>
+        <td style="border: 1px solid #ccc; padding: 10px; text-align: center;">(b,b)</td>
+        <td style="border: 1px solid #ccc; padding: 10px; text-align: center;">(b,b)</td>
+        <td style="border: 1px solid #ccc; padding: 10px; text-align: center;">(b,c)</td>
+        <td style="border: 1px solid #ccc; padding: 10px; text-align: center;">(b,c)</td>
+        <td style="border: 1px solid #ccc; padding: 10px; text-align: center;">(b,c)</td>
+      </tr>
+      <tr>
+        <th style="
+          border: 1px solid #ccc; 
+          padding: 10px; 
+          text-align: center; 
+          font-weight: bold; 
+          background-color: #f2f2f2;">Die 1: c</th>
+        <td style="border: 1px solid #ccc; padding: 10px; text-align: center;">(c,a)</td>
+        <td style="border: 1px solid #ccc; padding: 10px; text-align: center;">(c,b)</td>
+        <td style="border: 1px solid #ccc; padding: 10px; text-align: center;">(c,b)</td>
+        <td style="border: 1px solid #ccc; padding: 10px; text-align: center;">(c,c)</td>
+        <td style="border: 1px solid #ccc; padding: 10px; text-align: center;">(c,c)</td>
+        <td style="border: 1px solid #ccc; padding: 10px; text-align: center;">(c,c)</td>
+      </tr>
+    </tbody>
+  </table>
 
-$ Pr(A|B) = \frac{1}{11} $
+As we can see from the table, there are 11 possible outcomes that have at least 1 $a$ in them. Therefore, $Pr(B) = \frac{|B|}{|S|} = \frac{11}{36}$. <br />
+
+For $Pr(A \cap B)$, we only care about the cases when both rolls result in the same letter AND at least one of the rolls result in the letter $a$. <br />
+From the table, we can see that there is only 1 possible outcome that satisfies both conditions, which is the case where both dice roll an $a$. Therefore, $Pr(A \cap B) = \frac{|A \cap B|}{|S|} = \frac{1}{36}$. <br />
+
+Given this, we can now calculate $Pr(A|B)$ using the independence formula: <br />
+
+$ Pr(A|B) = \frac{ Pr(A \cap B) }{ Pr(B) } = \frac{ \frac{1}{36} }{ \frac{11}{36} } = \frac{1}{11} $ <br />