From cce0bf5b792b98627c963b789913058b92353775 Mon Sep 17 00:00:00 2001
From: John Lu <john.lu10212004@gmail.com>
Date: Thu, 18 Jul 2024 03:06:24 -0400
Subject: [PATCH] spaced out 2016 fall final

---
 .../comp2804/2016-fall-final/1/solution.md    |  5 ++
 .../comp2804/2016-fall-final/10/solution.md   | 15 ++++--
 .../comp2804/2016-fall-final/11/solution.md   | 12 +++--
 .../comp2804/2016-fall-final/12/solution.md   | 21 +++++----
 .../comp2804/2016-fall-final/13/solution.md   | 20 ++++----
 .../comp2804/2016-fall-final/16/solution.md   | 41 +++++++++--------
 .../comp2804/2016-fall-final/17/solution.md   | 46 ++++++++++---------
 .../comp2804/2016-fall-final/18/solution.md   |  6 ++-
 .../comp2804/2016-fall-final/19/solution.md   |  5 ++
 .../comp2804/2016-fall-final/2/solution.md    |  5 ++
 .../comp2804/2016-fall-final/20/solution.md   |  3 ++
 .../comp2804/2016-fall-final/21/solution.md   | 20 ++++----
 .../comp2804/2016-fall-final/22/solution.md   | 31 +++++++------
 .../comp2804/2016-fall-final/23/solution.md   | 28 ++++++-----
 .../comp2804/2016-fall-final/24/solution.md   |  6 +++
 .../comp2804/2016-fall-final/3/solution.md    | 29 ++++++------
 .../comp2804/2016-fall-final/4/solution.md    | 14 +++---
 .../comp2804/2016-fall-final/5/solution.md    | 32 ++++++++-----
 .../comp2804/2016-fall-final/7/solution.md    | 23 ++++++----
 .../comp2804/2016-fall-final/8/solution.md    | 27 ++++++-----
 20 files changed, 231 insertions(+), 158 deletions(-)

diff --git a/src/content/questions/comp2804/2016-fall-final/1/solution.md b/src/content/questions/comp2804/2016-fall-final/1/solution.md
index be737937..7ca264be 100644
--- a/src/content/questions/comp2804/2016-fall-final/1/solution.md
+++ b/src/content/questions/comp2804/2016-fall-final/1/solution.md
@@ -1,6 +1,11 @@
 Well, they aren't one-to-one functions, so each input has 7 possible outputs
+
 The first input has 9 possible outputs
+
 The second input has 9 possible outputs
+
 ...
+
 The seventh input has 9 possible outputs
+
 $ 9^7 $
diff --git a/src/content/questions/comp2804/2016-fall-final/10/solution.md b/src/content/questions/comp2804/2016-fall-final/10/solution.md
index aca0d080..8ca0aacb 100644
--- a/src/content/questions/comp2804/2016-fall-final/10/solution.md
+++ b/src/content/questions/comp2804/2016-fall-final/10/solution.md
@@ -1,6 +1,11 @@
-Well, the first character can be $ {a, b, c, d, e} $: 5 
-The second character can be $ {a, b, c, d, e} - { \text{first character} } $: 4 
-The third character can be $ {a, b, c, d, e} - { \text{second character} } $: 4 
-... 
-The nth character can be $ {a, b, c, d, e} - { \text{previous character} } $: 4 
+Well, the first character can be $ {a, b, c, d, e} $: 5
+
+The second character can be $ {a, b, c, d, e} - { \text{first character} } $: 4
+
+The third character can be $ {a, b, c, d, e} - { \text{second character} } $: 4
+
+...
+
+The nth character can be $ {a, b, c, d, e} - { \text{previous character} } $: 4
+
 $ 5 \cdot 4^{n-1} $
diff --git a/src/content/questions/comp2804/2016-fall-final/11/solution.md b/src/content/questions/comp2804/2016-fall-final/11/solution.md
index cec94eb9..0dc104d3 100644
--- a/src/content/questions/comp2804/2016-fall-final/11/solution.md
+++ b/src/content/questions/comp2804/2016-fall-final/11/solution.md
@@ -1,11 +1,13 @@
 We can use the stars and bars method to solve this problem.
+
 We have 4 dividers and 28 stars.
 
 <ul>
-<li> $x_1$ is the number of stars to the left of the first divider
-<li> $x_2$ is the number of stars between the first and second dividers
-<li> $x_3$ is the number of stars between the second and third dividers
-<li> $x_4$ is the number of stars between the third and fourth dividers
-<li> $x_5$ is the number of stars to the right of the fourth divider
+    <li> $x_1$ is the number of stars to the left of the first divider
+    <li> $x_2$ is the number of stars between the first and second dividers
+    <li> $x_3$ is the number of stars between the second and third dividers
+    <li> $x_4$ is the number of stars between the third and fourth dividers
+    <li> $x_5$ is the number of stars to the right of the fourth divider
 </ul>
+
 The number of solutions to the equation is $ \binom{28+4}{4} $
diff --git a/src/content/questions/comp2804/2016-fall-final/12/solution.md b/src/content/questions/comp2804/2016-fall-final/12/solution.md
index 03ac6f23..cf7a2f1b 100644
--- a/src/content/questions/comp2804/2016-fall-final/12/solution.md
+++ b/src/content/questions/comp2804/2016-fall-final/12/solution.md
@@ -1,13 +1,16 @@
 <ul>
-<li> Let's determine A <br/> 
-The possible outcomes are (1,5), (2,4), (3,3), (4,2), (5,1) <br/> 
-$ |A| = 5 $ <br/> 
-$ Pr(A) = \frac{5}{36} $
+	<li> Let's determine A <br/> 
+	The possible outcomes are (1,5), (2,4), (3,3), (4,2), (5,1) <br/> 
+	$ |A| = 5 $ <br/> 
+	$ Pr(A) = \frac{5}{36} $
 	<li> Let's determine $ A \cap B $ <br/> 
-	      The possible outcomes are (2,4) <br/> 
-	      $ |A \cap B| = 1 $ <br/> 
-	      $ Pr(A \cap B) = \frac{1}{36} $
+	The possible outcomes are (2,4) <br/> 
+	$ |A \cap B| = 1 $ <br/> 
+	$ Pr(A \cap B) = \frac{1}{36} $
 </ul>
-$ Pr(B|A) = \frac{Pr(A \cap B)}{Pr(A)} $ 
-$ Pr(B|A) = \frac{ \frac{1}{36}}{ \frac{5}{36}} $ 
+
+$ Pr(B|A) = \frac{Pr(A \cap B)}{Pr(A)} $
+
+$ Pr(B|A) = \frac{ \frac{1}{36}}{ \frac{5}{36}} $
+
 $ Pr(B|A) = \frac{1}{5} $
diff --git a/src/content/questions/comp2804/2016-fall-final/13/solution.md b/src/content/questions/comp2804/2016-fall-final/13/solution.md
index 55d85121..8f11a04d 100644
--- a/src/content/questions/comp2804/2016-fall-final/13/solution.md
+++ b/src/content/questions/comp2804/2016-fall-final/13/solution.md
@@ -1,12 +1,12 @@
 <ul>
-<li> Let's determine $ |V| $ <br/> 
-$ |V| = 20 $
-<li> Let's determine $ |W| $ <br/> 
-$ |W| = 7 $
-<li> Let's determine $ |A| $ <br/> 
-The number of ways to choose 4 even integers from 12 even integers: $ \binom{12}{4} $ <br/> 
-The number of ways to choose 3 odd integers from 8 odd integers: $ \binom{8}{3} $ <br/> 
-$ |A| = \binom{12}{4} \binom{8}{3} $
-<li> Let's determine $ Pr(A) $ <br/> 
-$ Pr(A) = \frac{\binom{12}{4} \binom{8}{3}}{\binom{20}{7}} $
+    <li> Let's determine $ |V| $ <br/> 
+    $ |V| = 20 $
+    <li> Let's determine $ |W| $ <br/> 
+    $ |W| = 7 $
+    <li> Let's determine $ |A| $ <br/> 
+    The number of ways to choose 4 even integers from 12 even integers: $ \binom{12}{4} $ <br/> 
+    The number of ways to choose 3 odd integers from 8 odd integers: $ \binom{8}{3} $ <br/> 
+    $ |A| = \binom{12}{4} \binom{8}{3} $
+    <li> Let's determine $ Pr(A) $ <br/> 
+    $ Pr(A) = \frac{\binom{12}{4} \binom{8}{3}}{\binom{20}{7}} $
 </ul>
diff --git a/src/content/questions/comp2804/2016-fall-final/16/solution.md b/src/content/questions/comp2804/2016-fall-final/16/solution.md
index a68d40ea..8d7ddc38 100644
--- a/src/content/questions/comp2804/2016-fall-final/16/solution.md
+++ b/src/content/questions/comp2804/2016-fall-final/16/solution.md
@@ -1,24 +1,27 @@
 <ul>
-<li> Let's determine $ S $ <br/> 
-S is the set of all possible outcomes of the birthdays of the students <br/> 
-$ |S| = 365^n $ <br/> 
-$ Pr(S) = \frac{365^n}{365^n} = 1 $
+	<li> Let's determine $ S $ <br/> 
+	S is the set of all possible outcomes of the birthdays of the students <br/> 
+	$ |S| = 365^n $ <br/> 
+	$ Pr(S) = \frac{365^n}{365^n} = 1 $
 	<li> Let's determine $ B $ <br/> 
-	      B is the set of all outcomes where no students have their birthday on December 14 <br/> 
-	      The first student has 364 possible birthdays <br/> 
-	      The second student has 364 possible birthdays <br/> 
-	      ... <br/> 
-	      The nth student has 364 possible birthdays <br/> 
-	      $ |B| = 364^n $ <br/> 
-	      $ Pr(B) = \frac{364^n}{365^n} $
+	B is the set of all outcomes where no students have their birthday on December 14 <br/> 
+	The first student has 364 possible birthdays <br/> 
+	The second student has 364 possible birthdays <br/> 
+	... <br/> 
+	The nth student has 364 possible birthdays <br/> 
+	$ |B| = 364^n $ <br/> 
+	$ Pr(B) = \frac{364^n}{365^n} $
 	<li> Let's determine $ C $ <br/> 
-	      C is the set of all outcomes where exactly one student has their birthday on December 14 <br/> 
-	      Out of n students, choose 1 of them to have their birthday on December 14: $ \binom{n}{1} = n $ <br/> 
-	      The student with the birthday on December 14 has 1 possible birthday: 1 <br/> 
-	      The other students have 364 possible birthdays: $ 364^{n-1} $ <br/> 
-	      $ |C| = n \cdot 1 \cdot 364^{n-1} $ <br/> 
-	      $ Pr(C) = \frac{n \cdot 1 \cdot 364^{n-1}}{365^n} $
+	C is the set of all outcomes where exactly one student has their birthday on December 14 <br/> 
+	Out of n students, choose 1 of them to have their birthday on December 14: $ \binom{n}{1} = n $ <br/> 
+	The student with the birthday on December 14 has 1 possible birthday: 1 <br/> 
+	The other students have 364 possible birthdays: $ 364^{n-1} $ <br/> 
+	$ |C| = n \cdot 1 \cdot 364^{n-1} $ <br/> 
+	$ Pr(C) = \frac{n \cdot 1 \cdot 364^{n-1}}{365^n} $
 </ul>
-Now, let's determine $ A $ 
-$ Pr(A) = Pr(S) - Pr(B) - Pr(C) $ 
+
+Now, let's determine $ A $
+
+$ Pr(A) = Pr(S) - Pr(B) - Pr(C) $
+
 $ Pr(A) = 1 - \frac{364^n}{365^n} - \frac{n \cdot 1 \cdot 364^{n-1}}{365^n} $
diff --git a/src/content/questions/comp2804/2016-fall-final/17/solution.md b/src/content/questions/comp2804/2016-fall-final/17/solution.md
index b130ab1b..e9003ece 100644
--- a/src/content/questions/comp2804/2016-fall-final/17/solution.md
+++ b/src/content/questions/comp2804/2016-fall-final/17/solution.md
@@ -1,27 +1,31 @@
 <ul>
-<li> Let's determine $ S $ <br/> 
-S is the set of all possible outcomes of the 17 students <br/> 
-$ |S| = \binom{100}{17} $
+	<li> Let's determine $ S $ <br/> 
+	S is the set of all possible outcomes of the 17 students <br/> 
+	$ |S| = \binom{100}{17} $
 	<li> Let's determine $ B $ <br/> 
-	      B occurs when 4 is in the subset <br/> 
-	      We choose 4: 1 <br/> 
-	      We choose 16 from the remaining 99: $ \binom{99}{16} $ <br/> 
-	      $ |B| = 1 \cdot \binom{99}{16} $ <br/> 
-	      $ Pr(B) = \frac{1 \cdot \binom{99}{16}}{\binom{100}{17}} $
+	B occurs when 4 is in the subset <br/> 
+	We choose 4: 1 <br/> 
+	We choose 16 from the remaining 99: $ \binom{99}{16} $ <br/> 
+	$ |B| = 1 \cdot \binom{99}{16} $ <br/> 
+	$ Pr(B) = \frac{1 \cdot \binom{99}{16}}{\binom{100}{17}} $
 	<li> Let's determine $ C $ <br/> 
-	      C occurs when 7 is in the subset <br/> 
-	      We choose 17: 1 <br/> 
-	      We choose 16 from the remaining 99: $ \binom{99}{16} $ <br/> 
-	      $ |C| = 1 \cdot \binom{99}{16} $ <br/> 
-	      $ Pr(C) = \frac{1 \cdot \binom{99}{16}}{\binom{100}{17}} $
+	C occurs when 7 is in the subset <br/> 
+	We choose 17: 1 <br/> 
+	We choose 16 from the remaining 99: $ \binom{99}{16} $ <br/> 
+	$ |C| = 1 \cdot \binom{99}{16} $ <br/> 
+	$ Pr(C) = \frac{1 \cdot \binom{99}{16}}{\binom{100}{17}} $
 	<li> Let's determine $ B \cap C $ <br/> 
-	      We choose 17: 1 <br/> 
-	      We choose 4: 1 <br/> 
-	      We choose 15 from the remaining 98: $ \binom{98}{15} $ <br/> 
-	      $ |B \cap C| = 1 \cdot 1 \cdot \binom{98}{15} $ <br/> 
-	      $ Pr(B \cap C) = \frac{1 \cdot 1 \cdot \binom{98}{15}}{\binom{100}{17}} $
+	We choose 17: 1 <br/> 
+	We choose 4: 1 <br/> 
+	We choose 15 from the remaining 98: $ \binom{98}{15} $ <br/> 
+	$ |B \cap C| = 1 \cdot 1 \cdot \binom{98}{15} $ <br/> 
+	$ Pr(B \cap C) = \frac{1 \cdot 1 \cdot \binom{98}{15}}{\binom{100}{17}} $
 </ul>
-Now, let's determine $ A $ 
-$ Pr(A) = Pr(B) + Pr(C) - Pr(B \cap C) $ 
-$ Pr(A) = \frac{1 \cdot \binom{99}{16}}{\binom{100}{17}} + \frac{1 \cdot \binom{99}{16}}{\binom{100}{17}} - \frac{1 \cdot 1 \cdot \binom{98}{15}}{\binom{100}{17}} $ 
+
+Now, let's determine $ A $
+
+$ Pr(A) = Pr(B) + Pr(C) - Pr(B \cap C) $
+
+$ Pr(A) = \frac{1 \cdot \binom{99}{16}}{\binom{100}{17}} + \frac{1 \cdot \binom{99}{16}}{\binom{100}{17}} - \frac{1 \cdot 1 \cdot \binom{98}{15}}{\binom{100}{17}} $
+
 $ Pr(A) = \frac{2 \cdot \binom{99}{16} - \binom{98}{15}}{\binom{100}{17}} $
diff --git a/src/content/questions/comp2804/2016-fall-final/18/solution.md b/src/content/questions/comp2804/2016-fall-final/18/solution.md
index 5b448d0d..de3cde10 100644
--- a/src/content/questions/comp2804/2016-fall-final/18/solution.md
+++ b/src/content/questions/comp2804/2016-fall-final/18/solution.md
@@ -1,5 +1,9 @@
 I'm not sure what the right way to do this is
+
 Forget the other values, let's just focus on 3,4, and 8
+
 There are 3! ways to arrange 3,4, and 8: $ {(3,4,8), (3,8,4), (4,3,8), (4,8,3), (8,3,4), (8,4,3) } $
-Out of those 6 ways, only 2 of them have 4 and 8 to the left of 3: $ { (4,8,3), (8,4,3) } $ 
+
+Out of those 6 ways, only 2 of them have 4 and 8 to the left of 3: $ { (4,8,3), (8,4,3) } $
+
 $ Pr(A) = \frac{2}{6} = \frac{1}{3} $
diff --git a/src/content/questions/comp2804/2016-fall-final/19/solution.md b/src/content/questions/comp2804/2016-fall-final/19/solution.md
index 28f09f03..51915f50 100644
--- a/src/content/questions/comp2804/2016-fall-final/19/solution.md
+++ b/src/content/questions/comp2804/2016-fall-final/19/solution.md
@@ -1,6 +1,11 @@
 Let $X_i$ be 1 if the next number is greater than the current number and 0 otherwise.
+
 The probability that a random number is greater than the previous number is $\frac{1}{2}$
+
 $ \mathbb{E}(X) = \mathbb{E}(X*1 + X_2 + \text{...} + X*{n-1}) $
+
 $ \mathbb{E}(X) = \mathbb{E}(X*1) + \mathbb{E}(X_2) + \text{...} + \mathbb{E}(X*{n-1}) $
+
 $ \mathbb{E}(X) = \frac{1}{2} + \frac{1}{2} + \text{...} + \frac{1}{2} $
+
 $ \mathbb{E}(X) = \frac{n-1}{2} $
diff --git a/src/content/questions/comp2804/2016-fall-final/2/solution.md b/src/content/questions/comp2804/2016-fall-final/2/solution.md
index 5f54b0c2..8eb2226b 100644
--- a/src/content/questions/comp2804/2016-fall-final/2/solution.md
+++ b/src/content/questions/comp2804/2016-fall-final/2/solution.md
@@ -1,6 +1,11 @@
 The first input has 9 possible outputs
+
 The second input has 8 possible outputs
+
 ...
+
 The seventh input has 3 possible outputs
+
 $ 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 $
+
 $ \frac{9!}{2!} $
diff --git a/src/content/questions/comp2804/2016-fall-final/20/solution.md b/src/content/questions/comp2804/2016-fall-final/20/solution.md
index 98748483..fac67537 100644
--- a/src/content/questions/comp2804/2016-fall-final/20/solution.md
+++ b/src/content/questions/comp2804/2016-fall-final/20/solution.md
@@ -1,4 +1,7 @@
 We take all possible unordered pairs by choosing 2 out of the n people: $ \binom{n}{2} $
+
 $P_i$'s birthday can be any day: $ \frac{365}{365} $
+
 $P_j$'s birthday must be the same day: $ \frac{1}{365} $
+
 $ \mathbb{E}(X) = \binom{n}{2} \cdot \frac{1}{365} $
diff --git a/src/content/questions/comp2804/2016-fall-final/21/solution.md b/src/content/questions/comp2804/2016-fall-final/21/solution.md
index 274709aa..220ca125 100644
--- a/src/content/questions/comp2804/2016-fall-final/21/solution.md
+++ b/src/content/questions/comp2804/2016-fall-final/21/solution.md
@@ -1,12 +1,16 @@
 <ul>
-<li> Let's determine X=5 <br/> 
-The probability of getting heads is $ \frac{1}{7} $ <br/> 
-$ Pr(X=5) = \frac{1}{7} $
+	<li> Let's determine X=5 <br/> 
+	The probability of getting heads is $ \frac{1}{7} $ <br/> 
+	$ Pr(X=5) = \frac{1}{7} $
 	<li> Let's determine X=2 <br/> 
-	      The probability of getting tails is $ \frac{6}{7} $ <br/> 
-	      $ Pr(X=2) = \frac{6}{7} $
+	The probability of getting tails is $ \frac{6}{7} $ <br/> 
+	$ Pr(X=2) = \frac{6}{7} $
 </ul>
-$ \mathbb{E}(X) = 5 \cdot Pr(X=5) + 2 \cdot Pr(X=2) $ 
-$ \mathbb{E}(X) = 5 \cdot \frac{1}{7} + 2 \cdot \frac{6}{7} $ 
-$ \mathbb{E}(X) = \frac{5}{7} + \frac{12}{7} $ 
+
+$ \mathbb{E}(X) = 5 \cdot Pr(X=5) + 2 \cdot Pr(X=2) $
+
+$ \mathbb{E}(X) = 5 \cdot \frac{1}{7} + 2 \cdot \frac{6}{7} $
+
+$ \mathbb{E}(X) = \frac{5}{7} + \frac{12}{7} $
+
 $ \mathbb{E}(X) = \frac{17}{7} $
diff --git a/src/content/questions/comp2804/2016-fall-final/22/solution.md b/src/content/questions/comp2804/2016-fall-final/22/solution.md
index c1ad4b1c..2079fd30 100644
--- a/src/content/questions/comp2804/2016-fall-final/22/solution.md
+++ b/src/content/questions/comp2804/2016-fall-final/22/solution.md
@@ -2,21 +2,24 @@ For questions like these, we can answer by checking for a value $i$ if the follo
 
 <ul>
 	<li> Let's determine $ Pr(X=1) $ <br/> 
-	      The coin that's heads has a probability of $ \frac{1}{2} $ <br/> 
-	      The other 6 coins have a probability of $ \frac{1}{2} $ <br/> 
-	      $ Pr(X=1) = { left( \frac{1}{2} \right)}^1 {\left( \frac{1}{2} \right) }^6 $ <br/> 
-	      $ Pr(X=1) = \frac{1}{2^7} $
+	The coin that's heads has a probability of $ \frac{1}{2} $ <br/> 
+	The other 6 coins have a probability of $ \frac{1}{2} $ <br/> 
+	$ Pr(X=1) = { left( \frac{1}{2} \right)}^1 {\left( \frac{1}{2} \right) }^6 $ <br/> 
+	$ Pr(X=1) = \frac{1}{2^7} $
 	<li> Let's determine $ Pr(Y=1) $ <br/> 
-	      The coin that's tails has a probability of $ \frac{1}{2} $ <br/> 
-	      The other 6 coins have a probability of $ \frac{1}{2} $ <br/> 
-	      $ Pr(Y=1) = { left( \frac{1}{2} \right) }^1 { left( \frac{1}{2} \right) }^6 $ <br/> 
-	      $ Pr(Y=1) = \frac{1}{2^7} $
+	The coin that's tails has a probability of $ \frac{1}{2} $ <br/> 
+	The other 6 coins have a probability of $ \frac{1}{2} $ <br/> 
+	$ Pr(Y=1) = { left( \frac{1}{2} \right) }^1 { left( \frac{1}{2} \right) }^6 $ <br/> 
+	$ Pr(Y=1) = \frac{1}{2^7} $
 	<li> Let's determine $ Pr(X=1 \cap Y=1) $ <br/> 
-	      If we get exactly 1 heads, then the remaining 6 coins must be tails <br/> 
-	      but that means we can't have exactly 1 tails as well <br/> 
-	      Since having both happen at the same time is impossible <br/> 
-	      $ Pr(X=1 \cap Y=1) = 0 $
+	If we get exactly 1 heads, then the remaining 6 coins must be tails <br/> 
+	but that means we can't have exactly 1 tails as well <br/> 
+	Since having both happen at the same time is impossible <br/> 
+	$ Pr(X=1 \cap Y=1) = 0 $
 </ul>
-$ Pr(X=i \cap Y=i) = Pr(X=i) \cdot Pr(Y=i) $ 
-$ 0 = \frac{1}{2^7} \cdot \frac{1}{2^7} $ 
+
+$ Pr(X=i \cap Y=i) = Pr(X=i) \cdot Pr(Y=i) $
+
+$ 0 = \frac{1}{2^7} \cdot \frac{1}{2^7} $
+
 Since the equation is false, X and Y are not independent.
diff --git a/src/content/questions/comp2804/2016-fall-final/23/solution.md b/src/content/questions/comp2804/2016-fall-final/23/solution.md
index ef8129ac..2fea6b28 100644
--- a/src/content/questions/comp2804/2016-fall-final/23/solution.md
+++ b/src/content/questions/comp2804/2016-fall-final/23/solution.md
@@ -2,19 +2,23 @@ For questions like these, we can answer by checking for a value $i$ if the follo
 
 <ul>
 	<li> Let's determine $ Pr(X=1) $ <br/> 
-	      Since half the numbers are odd, <br/> 
-	      $ Pr(X=1) = \frac{5}{10} $ <br/> 
-	      $ Pr(X=1) = \frac{1}{2} $
+	Since half the numbers are odd, <br/> 
+	$ Pr(X=1) = \frac{5}{10} $ <br/> 
+	$ Pr(X=1) = \frac{1}{2} $
 	<li> Let's determine $ Pr(Y=1) $ <br/> 
-	      Since 4 of the numbers are in the set {3, 4, 5, 6}, <br/> 
-	      $ Pr(Y=1) = \frac{4}{10} $ <br/> 
-	      $ Pr(Y=1) = \frac{2}{5} $
+	Since 4 of the numbers are in the set {3, 4, 5, 6}, <br/> 
+	$ Pr(Y=1) = \frac{4}{10} $ <br/> 
+	$ Pr(Y=1) = \frac{2}{5} $
 	<li> Let's determine $ Pr(X=1 \cap Y=1) $ <br/> 
-	      The values 3 and 5 are the only values that are both odd and in the set {3, 4, 5, 6} <br/> 
-	      $ Pr(X=1 \cap Y=1) = \frac{2}{10} $ <br/> 
-	      $ Pr(X=1 \cap Y=1) = \frac{1}{5} $
+	The values 3 and 5 are the only values that are both odd and in the set {3, 4, 5, 6} <br/> 
+	$ Pr(X=1 \cap Y=1) = \frac{2}{10} $ <br/> 
+	$ Pr(X=1 \cap Y=1) = \frac{1}{5} $
 </ul>
-$ Pr(X=1 \cap Y=1) = Pr(X=1) \cdot Pr(Y=1) $ 
-$ \frac{1}{5} = \frac{1}{2} \cdot \frac{2}{5} $ 
-$ \frac{1}{5} = \frac{1}{5} $ 
+
+$ Pr(X=1 \cap Y=1) = Pr(X=1) \cdot Pr(Y=1) $
+
+$ \frac{1}{5} = \frac{1}{2} \cdot \frac{2}{5} $
+
+$ \frac{1}{5} = \frac{1}{5} $
+
 Since the equation is true, X and Y are independent.
diff --git a/src/content/questions/comp2804/2016-fall-final/24/solution.md b/src/content/questions/comp2804/2016-fall-final/24/solution.md
index 4991c3f5..39af5f23 100644
--- a/src/content/questions/comp2804/2016-fall-final/24/solution.md
+++ b/src/content/questions/comp2804/2016-fall-final/24/solution.md
@@ -1,7 +1,13 @@
 Let's find the probability of rolling a number that is divisible by 3
+
 The numbers that are divisible by 3 are 3 and 6
+
 $ Pr(X=1) = \frac{2}{6} $
+
 We can use the geometric distribution to find the expected value of X
+
 Basically, just do $ \frac{1}{p} $
+
 $ \mathbb{E}(X) = \frac{1}{ \frac{2}{6}} $
+
 $ \mathbb{E}(X) = 3 $
diff --git a/src/content/questions/comp2804/2016-fall-final/3/solution.md b/src/content/questions/comp2804/2016-fall-final/3/solution.md
index 621fb669..40277a2a 100644
--- a/src/content/questions/comp2804/2016-fall-final/3/solution.md
+++ b/src/content/questions/comp2804/2016-fall-final/3/solution.md
@@ -1,18 +1,21 @@
 <ul>
-<li> Let A be the event that a bitstring of length 55 starts with 000. <br/> 
-The first 3 bits are fixed as 000. <br/> 
-The remaining 52 bits can be any combination of 0s and 1s: $2^{52}$. <br/> 
-$ |A| = 2^{52} $
+	<li> Let A be the event that a bitstring of length 55 starts with 000. <br/> 
+	The first 3 bits are fixed as 000. <br/> 
+	The remaining 52 bits can be any combination of 0s and 1s: $2^{52}$. <br/> 
+	$ |A| = 2^{52} $
 	<li> Let B be the event that a bitstring of length 55 ends with 1010. <br/> 
-	      The last 4 bits are fixed as 1010. <br/> 
-	      The first 51 bits can be any combination of 0s and 1s: $2^{51}$. <br/> 
-$ |B| = 2^{51} $
+	The last 4 bits are fixed as 1010. <br/> 
+	The first 51 bits can be any combination of 0s and 1s: $2^{51}$. <br/> 
+	$ |B| = 2^{51} $
 	<li> Let $ A \cap B $ be the event that a bitstring of length 55 starts with 000 AND ends with 1010. <br/> 
-	      The first 3 bits are fixed as 000. <br/> 
-	      The last 4 bits are fixed as 1010. <br/> 
-	      The remaining 48 bits can be any combination of 0s and 1s: $2^{48}$. <br/> 
-$ |A \cap B| = 2^{48} $
+	The first 3 bits are fixed as 000. <br/> 
+	The last 4 bits are fixed as 1010. <br/> 
+	The remaining 48 bits can be any combination of 0s and 1s: $2^{48}$. <br/> 
+	$ |A \cap B| = 2^{48} $
 </ul>
-Since it's asking for OR, we need to find $ A \cup B $ 
-$ |A \cup B| = |A| + |B| - |A \cap B| $ 
+
+Since it's asking for OR, we need to find $ A \cup B $
+
+$ |A \cup B| = |A| + |B| - |A \cap B| $
+
 $ |A \cup B| = 2^{52} + 2^{51} - 2^{48} $
diff --git a/src/content/questions/comp2804/2016-fall-final/4/solution.md b/src/content/questions/comp2804/2016-fall-final/4/solution.md
index 097ce953..ee1ebc0d 100644
--- a/src/content/questions/comp2804/2016-fall-final/4/solution.md
+++ b/src/content/questions/comp2804/2016-fall-final/4/solution.md
@@ -1,14 +1,14 @@
 Let's explain why some are wrong and some are right
 
 <ul>
-<li> The number of bitstrings of length $n$ that have at leats two 1's <br/> 
-If this was the case, it would subtract the set with all 0's and the set with a single 1 in any of the $n$ positions: $ 2^{n} - 1 - n $
+	<li> The number of bitstrings of length $n$ that have at leats two 1's <br/> 
+	If this was the case, it would subtract the set with all 0's and the set with a single 1 in any of the $n$ positions: $ 2^{n} - 1 - n $
 	<li> The number of bitstrings of length $n$ that have at most two 1's. <br/> 
-If this was the case, it would just be the set with all 0's and the set with a single 1 in any of the $n$ positions: $ 1 + n $
+	If this was the case, it would just be the set with all 0's and the set with a single 1 in any of the $n$ positions: $ 1 + n $
 	<li> The number of subsets of a set of size $n$ that have size at least two. <br/> 
-This one is saying \enquote{All subsets that are at least size two} <br/> 
-We subtract the empty set and choosing 1 element from the n elements: $ 2^{n} - 1 - n $
+	This one is saying \enquote{All subsets that are at least size two} <br/> 
+	We subtract the empty set and choosing 1 element from the n elements: $ 2^{n} - 1 - n $
 	<li> The number of subsets of a set of size $n$ that have size at least three. <br/> 
-This one is saying \enquote{All subsets that are at least size three} <br/> 
-We subtract the empty set, choosing 1 element from the n elements, and choosing 2 elements from the n elements: $ 2^{n} - 1 - n - \binom{n}{2} $
+	This one is saying \enquote{All subsets that are at least size three} <br/> 
+	We subtract the empty set, choosing 1 element from the n elements, and choosing 2 elements from the n elements: $ 2^{n} - 1 - n - \binom{n}{2} $
 </ul>
diff --git a/src/content/questions/comp2804/2016-fall-final/5/solution.md b/src/content/questions/comp2804/2016-fall-final/5/solution.md
index 743a1c5d..cadd8fad 100644
--- a/src/content/questions/comp2804/2016-fall-final/5/solution.md
+++ b/src/content/questions/comp2804/2016-fall-final/5/solution.md
@@ -1,18 +1,26 @@
 <ul>
-<li> Let S be the set of all students. <br/> 
-$ |S| = 100 $
+	<li> Let S be the set of all students. <br/> 
+	$ |S| = 100 $
 	<li> Let C be the event that someone drinks cider <br/> 
-	      $ |C| = 25 $
+	$ |C| = 25 $
 	<li> Let B be the event that someone drinks beer <br/> 
-	      $ |B| = 50 $
+	$ |B| = 50 $
 	<li> Let $ \overline{C} \cap \overline{B} $ be the event that someone does not drink cider and does not drink beer <br/> 
-	      $ | \overline{C} \cap \overline{B} | = 33 $
+	$ | \overline{C} \cap \overline{B} | = 33 $
 </ul>
-Find the union of C and B 
-$ |C \cup B| = |S| - | \overline{C} \cap \overline{B} | $ 
-$ |C \cup B| = 100 - 33 $ 
-$ |C \cup B| = 67 $ 
-Now, we find the intersection of C and B 
-$ |C \cup B| = |C| + |B| - |C \cap B| $ 
-$ 67 = 25 + 50 - |C \cap B| $ 
+
+Find the union of C and B
+
+$ |C \cup B| = |S| - | \overline{C} \cap \overline{B} | $
+
+$ |C \cup B| = 100 - 33 $
+
+$ |C \cup B| = 67 $
+
+Now, we find the intersection of C and B
+
+$ |C \cup B| = |C| + |B| - |C \cap B| $
+
+$ 67 = 25 + 50 - |C \cap B| $
+
 $ |C \cap B| = 8 $
diff --git a/src/content/questions/comp2804/2016-fall-final/7/solution.md b/src/content/questions/comp2804/2016-fall-final/7/solution.md
index 3eaaa520..af054cba 100644
--- a/src/content/questions/comp2804/2016-fall-final/7/solution.md
+++ b/src/content/questions/comp2804/2016-fall-final/7/solution.md
@@ -1,14 +1,17 @@
 <ul>
-<li> Let A be the event that there are two $a$'s <br/> 
-Out of n positions, choose 2 of them to be $a$'s: $ \binom{n}{2} $ <br/> 
-Everything else is a $b$, which is only 1 option. <br/> 
-$ |A| = \binom{n}{2} $
+	<li> Let A be the event that there are two $a$'s <br/> 
+	Out of n positions, choose 2 of them to be $a$'s: $ \binom{n}{2} $ <br/> 
+	Everything else is a $b$, which is only 1 option. <br/> 
+	$ |A| = \binom{n}{2} $
 	<li> Let B be the event that there are two $a$'s and they are next to each other <br/> 
-Out of n-1 positions, choose 1 of them to be the first $a$. <br/> 
-The next position is the second $a$. <br/> 
-Everything else is a $b$, which is only 1 option. <br/> 
-$ |B| = n-1 $
+	Out of n-1 positions, choose 1 of them to be the first $a$. <br/> 
+	The next position is the second $a$. <br/> 
+	Everything else is a $b$, which is only 1 option. <br/> 
+	$ |B| = n-1 $
 </ul>
-$S_n = |A| - |B|$ 
-$S_n = \binom{n}{2} - (n-1)$ 
+
+$S_n = |A| - |B|$
+
+$S_n = \binom{n}{2} - (n-1)$
+
 $S_n = \binom{n}{2} - n + 1$
diff --git a/src/content/questions/comp2804/2016-fall-final/8/solution.md b/src/content/questions/comp2804/2016-fall-final/8/solution.md
index 12cfae20..de73b968 100644
--- a/src/content/questions/comp2804/2016-fall-final/8/solution.md
+++ b/src/content/questions/comp2804/2016-fall-final/8/solution.md
@@ -1,19 +1,22 @@
 Let's calculate $f(1)$ and $f(2)$
+
 $ f(1) = \frac{5}{1} \cdot f(0) = 5 \cdot 1 = 5 $
+
 $ f(2) = \frac{5}{2} \cdot f(1) = \frac{5}{2} \cdot 5 = \frac{25}{2} $
 
 <ul>
-<li> $ f(n) = \frac{5}{n!} $ <br/> 
-$ f(1) = \frac{5}{1!} = 5 $ <br/> 
-$ f(2) = \frac{5}{2!} = \frac{5}{2} $
-<li> $ f(n) = \frac{5^{n}}{n!} $ <br/> 
-$ f(1) = \frac{5^{1}}{1!} = 5 $ <br/> 
-$ f(2) = \frac{5^{2}}{2!} = \frac{25}{2} $
-<li> $ f(n) = \frac{5^{n}}{(n+1)!} $ <br/> 
-$ f(1) = \frac{5^{1}}{2!} = \frac{5}{2} $ <br/> 
-$ f(2) = \frac{5^{2}}{3!} = \frac{25}{6} $
-<li> $ f(n) = \frac{5^{n+1}}{n!} $ <br/> 
-$ f(1) = \frac{5^{1+1}}{1!} = 25 $ <br/> 
-$ f(2) = \frac{5^{2+1}}{2!} = \frac{125}{2} $
+    <li> $ f(n) = \frac{5}{n!} $ <br/> 
+    $ f(1) = \frac{5}{1!} = 5 $ <br/> 
+    $ f(2) = \frac{5}{2!} = \frac{5}{2} $
+    <li> $ f(n) = \frac{5^{n}}{n!} $ <br/> 
+    $ f(1) = \frac{5^{1}}{1!} = 5 $ <br/> 
+    $ f(2) = \frac{5^{2}}{2!} = \frac{25}{2} $
+    <li> $ f(n) = \frac{5^{n}}{(n+1)!} $ <br/> 
+    $ f(1) = \frac{5^{1}}{2!} = \frac{5}{2} $ <br/> 
+    $ f(2) = \frac{5^{2}}{3!} = \frac{25}{6} $
+    <li> $ f(n) = \frac{5^{n+1}}{n!} $ <br/> 
+    $ f(1) = \frac{5^{1+1}}{1!} = 25 $ <br/> 
+    $ f(2) = \frac{5^{2+1}}{2!} = \frac{125}{2} $
 </ul>
+
 Since b is the only one that has the correct function, it is the answer