added 2019 winter midterm

src/content/questions/comp2804/2019-fall-final/1/solution.md

@@ -1,3 +1,5 @@
 We choose 5 of the 85 positions to be c: $ \binom{85}{5} $
+
 The remaining 80 positions can be any of the 3 letters: $ 3^{80} $
+
 Thus, the total number of strings is $ \binom{85}{5} \cdot 3^{80} $

src/content/questions/comp2804/2019-fall-final/10/solution.md

@@ -1,3 +1,5 @@
 Let's just draw the tree, man
+
 Warning: Do Later
+
 In total, O Canada is sung 8 times.

src/content/questions/comp2804/2019-fall-final/11/solution.md

@@ -6,6 +6,9 @@
     The remaining 8 each have 3 possible incorrect answers: $ 3^{8} $ <br/> 
     $ |A| = \binom{25}{17} \cdot 3^{8} $
 </ul>
-$ Pr(A) = \frac{|A|}{|S|} $ 
+
+$ Pr(A) = \frac{|A|}{|S|} $
+
 $ Pr(A) = \frac{\binom{25}{17} \cdot 3^{8}}{4^{25}} $
+
 $ Pr(A) = \binom{25}{17} \cdot {( \frac{3}{4})}^8 \cdot {( \frac{1}{4})}^{17} $

src/content/questions/comp2804/2019-fall-final/12/solution.md

@@ -19,8 +19,13 @@ I've been playing an MMO RPG called Orna recently. It uses GPS and forces me to
     $ |A \cap B| = 1 $ <br/> 
     $ Pr(A \cap B) = \frac{1}{32} $
 </ul>
-Now, let's check whether it's independent 
-$ Pr(A \cap B) = Pr(A) \cdot Pr(B) $ 
-$ \frac{1}{32} = \frac{1}{4} \cdot \frac{1}{8} $ 
-$ \frac{1}{32} = \frac{1}{32} $ 
+
+Now, let's check whether it's independent
+
+$ Pr(A \cap B) = Pr(A) \cdot Pr(B) $
+
+$ \frac{1}{32} = \frac{1}{4} \cdot \frac{1}{8} $
+
+$ \frac{1}{32} = \frac{1}{32} $
+
 BOOM. THE RESULTS SPEAK FOR THEMSELVES. HAPPY INDEPENDENCE DAY. MURICAAAA

src/content/questions/comp2804/2019-fall-final/15/solution.md

@@ -7,6 +7,9 @@
     We can choose 2 of the red balls: $ \binom{4}{2} $ <br/> 
     $ |A \cap B| = \binom{4}{2} = 6 $
 </ul>
+
 Now, let God do the rest
-$ Pr(A|B) = \frac{|A \cap B|}{|B|} = $ 
+
+$ Pr(A|B) = \frac{|A \cap B|}{|B|} = $
+
 $ Pr(A|B) = \frac{ \binom{5}{2} + \binom{4}{2} }{ \binom{4}{2} } $

src/content/questions/comp2804/2019-fall-final/16/solution.md

@@ -6,5 +6,7 @@
     There is only 1 number that is both even and divisible by 3, which is 6: 1 <br/> 
     $ |A \cap B| = 1 $
 </ul>
-I enjoy Jujutsu Kaisen and Donal/Joe/Obama AI voice skits 
+
+I enjoy Jujutsu Kaisen and Donal/Joe/Obama AI voice skits
+
 $ Pr(A|B) = \frac{|A \cap B|}{|B|} = \frac{1}{3} $

src/content/questions/comp2804/2019-fall-final/17/solution.md

@@ -14,9 +14,15 @@
     $ |A \cap B| = 2 $ <br/> 
     $ Pr(A \cap B) = \frac{2}{10} = \frac{1}{5} $
 </ul>
-Avatar: The Last Airbender is pretty high tier 
-Let's check for independence 
-$ Pr(A \cap B) = Pr(A) \cdot Pr(B) $ 
-$ \frac{1}{5} = \frac{1}{2} \cdot \frac{1}{2} $ 
-$ \frac{1}{5} = \frac{1}{4} $ 
+
+Avatar: The Last Airbender is pretty high tier
+
+Let's check for independence
+
+$ Pr(A \cap B) = Pr(A) \cdot Pr(B) $
+
+$ \frac{1}{5} = \frac{1}{2} \cdot \frac{1}{2} $
+
+$ \frac{1}{5} = \frac{1}{4} $
+
 The equation is not false; therefore, they're dependent

src/content/questions/comp2804/2019-fall-final/18/solution.md

@@ -14,9 +14,15 @@
     $ |A \cap B| = 3 $ <br/> 
     $ Pr(A \cap B) = \frac{3}{10} $
 </ul>
-Caedrel is the best rat 
-Let's check for independence 
-$ Pr(A \cap B) = Pr(A) \cdot Pr(B) $ 
-$ \frac{3}{10} = \frac{1}{2} \cdot \frac{3}{5} $ 
-$ \frac{3}{10} = \frac{3}{10} $ 
+
+Caedrel is the best rat
+
+Let's check for independence
+
+$ Pr(A \cap B) = Pr(A) \cdot Pr(B) $
+
+$ \frac{3}{10} = \frac{1}{2} \cdot \frac{3}{5} $
+
+$ \frac{3}{10} = \frac{3}{10} $
+
 Therefore, it is independent

src/content/questions/comp2804/2019-fall-final/2/solution.md

@@ -1,4 +1,7 @@
 We choose 15 of the 85 positions to be a: $ \binom{85}{15} $
+
 We choose 30 of the 70 remaining positions to be d: $ \binom{70}{30} $
+
 The remaining 40 positions can be any of the 2 letters: $ 2^{40} $
+
 Thus, the total number of strings is $ \binom{85}{15} \cdot \binom{70}{30} \cdot 2^{40} $

src/content/questions/comp2804/2019-fall-final/20/solution.md

@@ -1,10 +1,19 @@
 Let $X_1$ be the amount of dollars you win if the first coin comes up tails: $-1$
+
 There's a 50/50 chance of it landing tails
+
 $ Pr(X_1 = -1) = \frac{1}{2} $
+
 Let $X_2$ be the amount of dollars you win if the second coin comes up heads: $2$
+
 There's a 50/50 chance of it landing heads
+
 $ Pr(X_2 = 2) = \frac{1}{2} $
+
 $ E(X) = (-1) \cdot Pr(X_1 = -1) + 2 \cdot Pr(X_2 = 2) $
+
 $ E(X) = (-1) \cdot \frac{1}{2} + 2 \cdot \frac{1}{2} $
+
 $ E(X) = -\frac{1}{2} + 1 $
+
 $ E(X) = \frac{1}{2} $

src/content/questions/comp2804/2019-fall-final/24/solution.md

@@ -19,11 +19,19 @@ Let $X_i$ be 1 if the students at positions $i$ and $2i$ are politically correct
     $ Pr (X_i=1 ) = \frac{ 1 }{ 1 } \cdot \frac{k \cdot (k-1) \cdot 1 }{n \cdot (n-1) } $ <br/> 
     $ Pr (X_i=1 ) = \frac{k \cdot (k-1) }{n \cdot (n-1) } $
 </ul>
-Well, $S_n$ corresponds to $S*{ \frac{n}{2} }$ 
-$S_{n-2}$ corresponds to $S_{ \frac{n}{2}-1}$ 
-$S_{n-4}$ corresponds to $S_{ \frac{n}{2}-2}$ 
-Half of everyone has a corresponding junior 
-$ E(X) = \sum*{i=1}^{n/2} Pr(X_i=1) $ 
-$ E(X) = \sum*{i=1}^{n/2} \frac{k \cdot (k-1) }{n \cdot (n-1) } $ 
-$ E(X) = \frac{k (k-1) }{n (n-1) } \cdot \frac{n}{2} $ 
+
+Well, $S_n$ corresponds to $S*{ \frac{n}{2} }$
+
+$S_{n-2}$ corresponds to $S_{ \frac{n}{2}-1}$
+
+$S_{n-4}$ corresponds to $S_{ \frac{n}{2}-2}$
+
+Half of everyone has a corresponding junior
+
+$ E(X) = \sum\*{i=1}^{n/2} Pr(X_i=1) $
+
+$ E(X) = \sum\*{i=1}^{n/2} \frac{k \cdot (k-1) }{n \cdot (n-1) } $
+
+$ E(X) = \frac{k (k-1) }{n (n-1) } \cdot \frac{n}{2} $
+
 IT'S OVER. THE PAIN. THE SUFFERING. THE NIGHTMARES.

src/content/questions/comp2804/2019-fall-final/7/solution.md

@@ -1,6 +1,9 @@
 Yeah, so this is the dividers method. Let's define some stuff I guess
+
 Let $x_1, x_2, x_3, x_4, x_5, x_6, x_7$ be the number of bananas Nick eats on each day
+
 Now, $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 = 25$
+
 Assuming you have a sea of bananas, let's put dividers to split them off into different days
 
 <ul>

src/content/questions/comp2804/2019-fall-final/9/solution.md

@@ -1,5 +1,9 @@
 Warning: I suck at these questions. Here goes nothing
+
 Possibilities:
+
 $0, S_{n-1}$
+
 $1,0, S_{n-2}$
+
 IDK MAN

src/content/questions/comp2804/2019-winter-midterm/1/solution.md

@@ -1,3 +1,5 @@
 There are 2 choices for the odd positions: 0 or 1.
+
 There are 6 even positions that can each be 0 or 1, so there are $ 2^6 $ ways to choose the characters for the even positions.
+
 Thus, there are $ 2 \cdot 2^6 =2 \cdot 64 = 128 $ such bitstrings.

src/content/questions/comp2804/2019-winter-midterm/10/solution.md

@@ -1,5 +1,9 @@
 Because there is a 0 at position 59, there must be a 1 at position 58 and 60 to avoid a 00.
+
 It looks something like this: ..., 1, 0, 1, ..., 1, ...
-The number of 00-free bitstrings made from the first 57 bits $ \text{bit positions 1 to 57} $ is $ f*{57+2} $.
-The number of 00-free bitstrings made from the 17 bits $ \text{bit positions 61 to 77} $ is $ f*{17+2} $.
+
+The number of 00-free bitstrings made from the first 57 bits $ \text{bit positions 1 to 57} $ is $ f\*{57+2} $.
+
+The number of 00-free bitstrings made from the 17 bits $ \text{bit positions 61 to 77} $ is $ f\*{17+2} $.
+
 Thus, the number of 00-free bitstrings with the above conditions is $ f*{59} \cdot f*{19} $.

src/content/questions/comp2804/2019-winter-midterm/14/solution.md

@@ -1,2 +1,3 @@
 If $ (a) $ is true, then it would not be recursive
+
 $ (b) $ is pretty much what a recurisve function is

src/content/questions/comp2804/2019-winter-midterm/15/solution.md

@@ -3,12 +3,21 @@
     <li> Guinness Extra Stout,
     <li> Magners Original Irish Cider.
 </ul>
-There is an unlimited supply for each of these types. 
-There are 75 students at the party, and each of them gets one drink, which is chosen uniformly at random from these three types. 
+
+There is an unlimited supply for each of these types.
+
+There are 75 students at the party, and each of them gets one drink, which is chosen uniformly at random from these three types.
+
 Let be the event $ A = \text{\enquote{exactly 50 students get Magners Original Irish Cider}.} $
+
 What is Pr $ (A) $?
-We need to choose 50 students to get Magners Original Irish Cider from the 75 students: $ \binom{75}{50} $ 
-For each of the 50 students, there is a $ \frac{1}{3} $ chance of getting Magners Original Irish Cider: $ {( \frac{1}{3})}^{50} $ 
-For each of the 25 students, there is a $ \frac{2}{3} $ chance of not getting Magners Original Irish Cider: $ {( \frac{2}{3})}^{25} $ 
-Pr $ (A) = \binom{75}{50} \cdot {( \frac{1}{3})}^{50} \cdot {( \frac{2}{3})}^{25} $ 
+
+We need to choose 50 students to get Magners Original Irish Cider from the 75 students: $ \binom{75}{50} $
+
+For each of the 50 students, there is a $ \frac{1}{3} $ chance of getting Magners Original Irish Cider: $ {( \frac{1}{3})}^{50} $
+
+For each of the 25 students, there is a $ \frac{2}{3} $ chance of not getting Magners Original Irish Cider: $ {( \frac{2}{3})}^{25} $
+
+Pr $ (A) = \binom{75}{50} \cdot {( \frac{1}{3})}^{50} \cdot {( \frac{2}{3})}^{25} $
+
 Pr $ (A) = \frac{ \binom{75}{50} \cdot 2^{25}}{3^{75}} $

src/content/questions/comp2804/2019-winter-midterm/2/solution.md

@@ -1,9 +1,17 @@
 A = Permutations that contain $ bge $
+
 We can treat $ bge $ as a single entity.
+
 B = Permutations that do not contain $ bge $
+
 So there's $ bge, a,c,d,f $ which is 5 entities.
+
 $ |B| = 5! $
+
 $ |S| = $ Total permutations without restrictions
+
 $ |S| = 7! $
+
 $ |\text{Number of permutations that do not contain} bge| = |S| - |B| $
+
 $ |\text{Number of permutations that do not contain} bge| = 7! - 5! $

src/content/questions/comp2804/2019-winter-midterm/3/solution.md

@@ -1,9 +1,17 @@
 A = Strings that contain at least one lowercase letter
+
 $ \overline{A} = $ Strings that contain no lowercase letters
+
 $ |\overline{A}| = 26^{15} $
+
 B = Strings that contain at least one uppercase letter
+
 $ \overline{B} = $ Strings that contain no uppercase letters
+
 $ |\overline{B}| = 26^{15} $
+
 $ |A \cap B| = \text{All Possibilities} - |\overline{A}| - |\overline{B}| $
+
 $ |A \cap B| = 52^{15} - 26^{15} - 26^{15} $
+
 $ |A \cap B| = 52^{15} - 2 \cdot 26^{15} $

src/content/questions/comp2804/2019-winter-midterm/4/solution.md

@@ -1,2 +1,3 @@
 $ \text{All Possibilities} - \text{0 even numbers} - \text{1 even number} $
+
 $ \binom{n}{k} - \binom{n/2}{k} - \binom{n/2}{k-1} $

src/content/questions/comp2804/2019-winter-midterm/5/solution.md

@@ -1,10 +1,19 @@
 A = Like beer
+
 $ |A| = 37 $
+
 B = Like cider
+
 $ |B| = 18 $
+
 $ |\overline{A \cup B}| = 55 $
+
 $ |A \cup B| = 100-55 $
+
 $ |A \cup B| = 45 $
+
 $ |A \cap B| = |A| + |B| - |A \cup B| $
+
 $ |A \cap B| = 37 + 18 - 45 $
+
 $ |A \cap B| = 10 $

src/content/questions/comp2804/2019-winter-midterm/6/solution.md

@@ -1,6 +1,11 @@
 We can use the pigeonhole principle to solve this problem.
+
 We need to have 1 more than the maximum number of possible grades.
+
 That way, we can guarantee that if all students receive different grades, the last student will receive a duplicate grade.
+
 There are 5 possible grades: $ A, B, C, D, F $
+
 Thus, the minimum value for $ n $ is $ 5 \cdot 3 + 1 $
+
 Thus, the minimum value for $ n $ is 16.

src/content/questions/comp2804/2019-winter-midterm/7/solution.md

@@ -1,2 +1,3 @@
 We can treat the largest element as a fixed element.
+
 We need to choose 16 other elements from the remaining 29 elements that are smaller than 30: $ \binom{29}{16} $

src/content/questions/comp2804/2019-winter-midterm/9/solution.md

@@ -1,4 +1,7 @@
 There are $ n/2 $ even positions.
+
 We need to choose $ k $ of these positions to be $ a $'s: $ \binom{n/2}{k} $
+
 The remaining $ n - k $ positions must be $ b $'s or $ c $'s: $ 2^{n-k} $
+
 Thus, there are $ \binom{n/2}{k} \cdot 2^{n-k} $ such strings.