spaced out 2017 winter final

src/content/questions/comp2804/2017-winter-final/1/solution.md

@@ -12,4 +12,5 @@ Well, let's break it down
     We can place the last even number in the last remaining even position: 1 <br/> 
     This creates $ 5! $ permutations
 </ul>
+
 Multiplying the two permutations, we get $ 5! \cdot 5! $

src/content/questions/comp2804/2017-winter-final/10/solution.md

@@ -1,10 +1,19 @@
 Let S be all possible strings of length 14: $ |S| = 26^{14} $
+
 Let B be the event that the string contains at no vowels
+
 The first character can be any of the 21 consonants: 21
+
 The second character can be any of the 21 consonants: 21
+
 ...
+
 The 14th character can be any of the 21 consonants: 21
+
 $ |B| = 21^{14} $
+
 $ Pr(B) = \frac{21^{14}}{26^{14}} $
+
 $ Pr(A) = 1 - Pr(B) $
+
 $ Pr(A) = 1 - \frac{21^{14}}{26^{14}} $

src/content/questions/comp2804/2017-winter-final/11/solution.md

@@ -1,11 +1,21 @@
 This means there are 13 positions
+
 First, select 2 of the girls to be Elisa's neighbors: $ \binom{6}{2} $
+
 Since we're putting 2 girls beside Elisa, girl 1 can be on the left and girl 2 can be on the right OR girl 1 can be on the right and girl 2 can be on the left: 2
+
 As a single entity, the 3 girls can be placed, starting from position 0 to position 10: 11
+
 The remaining 10 people can be placed in the remaining 10 positions: 10!
+
 $ \binom{6}{2} \cdot 2 \cdot 11 \cdot 10! $
+
 $ = \frac{6!}{2!4!} \cdot 2 \cdot 11 \cdot 10! $
+
 $ = \frac{6!}{4!} \cdot 11 \cdot 10! $
+
 $ = \frac{6 \cdot 5 \cdot 4!}{4!} \cdot 11 \cdot 10! $
+
 $ = \frac{6 \cdot 5}{1} \cdot 11 \cdot 10! $
+
 $ = 6 \cdot 5 \cdot 11 \cdot 10! $

src/content/questions/comp2804/2017-winter-final/12/solution.md

@@ -17,7 +17,11 @@
     $ |A \cap B| = \binom{8}{3} $ <br/> 
     $ Pr(A \cap B) = \frac{ \binom{8}{3} }{ \binom{10}{5} } = \frac{8!}{5!3!} div \frac{10!}{5!5!} = \frac{8!}{5!3!} \cdot \frac{5!5!}{10!} = \frac{8!}{3!} \cdot \frac{5!}{10!} = \frac{1}{3!} \cdot \frac{5!}{10 \cdot 9} = \frac{1}{1} \cdot \frac{5 \cdot 4}{10 \cdot 9} = \frac{5 \cdot 2}{5 \cdot 9} = \frac{2}{9} $
 </ul>
-$ Pr(A|B) = \frac{ Pr(A \cap B) }{ Pr(B) } $ 
-$ Pr(A|B) = \frac{ \frac{2}{9} }{ \frac{1}{2} } $ 
-$ Pr(A|B) = \frac{2}{9} \cdot \frac{2}{1} $ 
+
+$ Pr(A|B) = \frac{ Pr(A \cap B) }{ Pr(B) } $
+
+$ Pr(A|B) = \frac{ \frac{2}{9} }{ \frac{1}{2} } $
+
+$ Pr(A|B) = \frac{2}{9} \cdot \frac{2}{1} $
+
 $ Pr(A|B) = \frac{4}{9} $

src/content/questions/comp2804/2017-winter-final/13/solution.md

@@ -10,7 +10,11 @@
     $ | B \cap A | = 4 $ <br/> 
     $ Pr(A \cap B) = \frac{4}{32} = \frac{1}{8} $
 </ul>
-$ Pr(A|B) = \frac{ Pr(A \cap B) }{ Pr(B) } $ 
-$ Pr(A|B) = \frac{ \frac{1}{8} }{ \frac{1}{2} } $ 
-$ Pr(A|B) = \frac{1}{8} \cdot \frac{2}{1} $ 
+
+$ Pr(A|B) = \frac{ Pr(A \cap B) }{ Pr(B) } $
+
+$ Pr(A|B) = \frac{ \frac{1}{8} }{ \frac{1}{2} } $
+
+$ Pr(A|B) = \frac{1}{8} \cdot \frac{2}{1} $
+
 $ Pr(A|B) = \frac{1}{4} $

src/content/questions/comp2804/2017-winter-final/14/solution.md

@@ -21,8 +21,13 @@
     $ |A \cap B| = \frac{(n-1)!}{2!} $ <br/> 
     $ Pr(A \cap B) = \frac{(n-1)!}{2 \cdot n!} = \frac{1}{2n} $
 </ul>
-Now, let's check for independence 
-$ Pr(A \cap B) = Pr(A) \cdot Pr(B) $ 
-$ \frac{1}{2n} = \frac{1}{n} \cdot \frac{1}{2} $ 
-$ \frac{1}{2n} = \frac{1}{2n} $ 
+
+Now, let's check for independence
+
+$ Pr(A \cap B) = Pr(A) \cdot Pr(B) $
+
+$ \frac{1}{2n} = \frac{1}{n} \cdot \frac{1}{2} $
+
+$ \frac{1}{2n} = \frac{1}{2n} $
+
 Since the equation is true, A and B are independent

src/content/questions/comp2804/2017-winter-final/15/solution.md

@@ -9,6 +9,9 @@
     The probability that both $a_1 = 1$ and $a_n = 5$ is $ \frac{1}{n^2} $ <br/> 
     $ Pr(A \cap B) = \frac{1}{n^2} $
 </ul>
-$ Pr(A \cup B) = Pr(A) + Pr(B) - Pr(A \cap B) $ 
-$ Pr(A \cup B) = \frac{1}{n} + \frac{1}{n} - \frac{1}{n} \frac{1}{n-1} $ 
+
+$ Pr(A \cup B) = Pr(A) + Pr(B) - Pr(A \cap B) $
+
+$ Pr(A \cup B) = \frac{1}{n} + \frac{1}{n} - \frac{1}{n} \frac{1}{n-1} $
+
 $ Pr(A \cup B) = \frac{2}{n} - \frac{1}{n(n-1)} $

src/content/questions/comp2804/2017-winter-final/16/solution.md

@@ -1,14 +1,27 @@
 Let's make some statements and see what we can get
+
 $ Pr(A|B) = \frac{Pr(A \cap B)}{Pr(B)} $
+
 $ Pr(B|A) = \frac{Pr(A \cap B)}{Pr(A)} $
+
 $ \frac{Pr(A \cap B)}{Pr(A)} = \frac{Pr(A \cap B)}{Pr(B)} $
+
 $ Pr(A) = Pr(B) $
+
 $ Pr(A \cup B) = Pr(A) + Pr(B) - Pr(A \cap B) = 1 $
+
 $ 2Pr(A) - Pr(A \cap B) = 1 $
+
 For the above statement to be true, $ Pr(A) > \frac{1}{2} $ because $ Pr(A \cap B) > 0 $
+
 To illustrate, let's use some values
+
 $ Pr(A) = 0.5 $ and $ Pr(A \cap B) = 0.25 $
+
 $ 2 \cdot 0.5 - 0.25 = 1 $
+
 $ 1 - 0.25 = 1 $
+
 $ 0.75 = 1 $
+
 As can be seen, $ Pr(A) > \frac{1}{2} $ needs to be true in order to counter the fact that $ Pr(A \cap B) > 0 $

src/content/questions/comp2804/2017-winter-final/17/solution.md

@@ -1,5 +1,9 @@
 For all intents and purposes, the set we really care about is $ |S| = 4! $
+
 We can do $ { (1,2), (2,1) } $ on the left side: 2
+
 We can do $ { (8,4), (4,8) } $ on the right side: 2
+
 $ |A| = 2 \cdot 2 $
+
 $ Pr(A) = \frac{4}{24} = \frac{1}{6} $

src/content/questions/comp2804/2017-winter-final/19/solution.md

@@ -1,7 +1,13 @@
 Let's check for the probability that 3 people have the same birthday for every subset of 3 people
+
 We choose 3 people out of the n people: $ \binom{n}{3} $
+
 The first dude in the subset can have birthday on any day of the year: $ \frac{365}{365} $
+
 The second bro in the subset must have the same birthday as the first dude: $ \frac{1}{365} $
+
 The thid mate in the subset must have the same birthday as the first dude: $ \frac{1}{365} $
+
 The probability that 3 people have the same birthday is $ \frac{1}{365^2} $
+
 $ \mathbb{E}(X) = \binom{n}{3} \cdot \frac{1}{365^2} $

src/content/questions/comp2804/2017-winter-final/2/solution.md

@@ -1,6 +1,11 @@
 Well, let's break it down
+
 We choose 2 numbers out of the n numbers: $ \binom{n}{2} $
+
 The other n-2 numbers can be placed in any order: $ (n-2)! $
+
 $ \binom{n}{2} \cdot (n-2)! $
+
 $ = \frac{n!}{2! \cdot (n-2)!} \cdot (n-2)! $
+
 $ = \frac{n!}{2!} $

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

@@ -1,8 +1,15 @@
 Let's calculate the expected value of the each flip
+
 $ \mathbb{E}(X) = 100 \cdot Pr(\text{Heads}) + 50 \cdot Pr(\text{Tails}) $
+
 $ 100 \cdot \frac{1}{5} + 50 \cdot \frac{4}{5} $
+
 $ = 20 + 40 $
+
 $ = 60 $
-Since we flip 2 coins, we double the expected value 
+
+Since we flip 2 coins, we double the expected value
+
 $ \mathbb{E}(X) = 60 \cdot 2 $
+
 $ \mathbb{E}(X) = 120 $

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

@@ -11,8 +11,13 @@ We don't need to calculate all possibilities. Since this is independence, we're
     If the only rolls that satisfy condition 1 are $ (1,6), (6,1) $ and the only rolls that satisfy condition 2 are $ (1,1) $, then there are no rolls that satisfy both conditions <br/> 
     $ Pr(X=5 \cap Y=1) = 0 $
 </ul>
-Now, check time 
-$ Pr(X=5 \cap Y=1) = Pr(X=5) \cdot Pr(Y=1) $ 
-$ 0 = \frac{1}{18} \cdot \frac{1}{36} $ 
-$ 0 = \frac{1}{648} $ 
+
+Now, check time
+
+$ Pr(X=5 \cap Y=1) = Pr(X=5) \cdot Pr(Y=1) $
+
+$ 0 = \frac{1}{18} \cdot \frac{1}{36} $
+
+$ 0 = \frac{1}{648} $
+
 Since the equation is false, there are dependent

src/content/questions/comp2804/2017-winter-final/22/solution.md

@@ -1,5 +1,9 @@
 Idk, I'll just write down what I see and try to find a pattern
+
 We have a 1 in 4 chance of landing 2 heads per double coin flip
+
 $ Pr(X = 1) + Pr(X = 2) + Pr(X = 3) + Pr(X = 4) + ... $
+
 $ \frac{1}{4} + ( \frac{3}{4})( \frac{1}{4}) + {( \frac{3}{4})}^2 ( \frac{1}{4}) + ... $
+
 From this, I got the formula $ {( \frac{3}{4})}^{m-1} \cdot \frac{1}{4} $

src/content/questions/comp2804/2017-winter-final/23/solution.md

@@ -1,2 +1,3 @@
 The answer is that it's cap. Linearity of expectations only works on addition and subtraction.
+
 I call \cap on multiplication and division

src/content/questions/comp2804/2017-winter-final/24/solution.md

@@ -1,9 +1,17 @@
 Thank God I have a second monitor to watch YouTube videos on
+
 Alright, so let's break it down. It's asking for the probability that a $D_i$ is a cider and $D_{i+1}$ is not a cider
+
 First has probability of being cider $ \frac{11}{12} $
+
 Second has probability of not being cider $ \frac{1}{12} $
+
 This occurs 6 times. Not 7 because $D_n$ has no $D_{n+1}$ to compare it to
+
 $ \mathbb{E}(X) = \sum\_{i=1}^{6} \frac{11}{12} \cdot \frac{1}{12} $
+
 $ \mathbb{E}(X) = 6 \cdot \frac{11}{12} \cdot \frac{1}{12} $
+
 $ \mathbb{E}(X) = 6 \cdot \frac{11}{144} $
+
 $ \mathbb{E}(X) = \frac{6}{144} $

src/content/questions/comp2804/2017-winter-final/3/solution.md

@@ -1,4 +1,5 @@
 Holy moly, that's a lot of words
+
 Let's do this anyway$ ( $cries internally $ ) $
 
 <ul>
@@ -16,5 +17,7 @@ Let's do this anyway$ ( $cries internally $ ) $
     We have exactly 5 beer bottles and 5 cider bottles <br/> 
     $ | B \cap C | = \binom{17}{5} \cdot \binom{28}{5} $
 </ul>
-$ B \cup C = |B| + |C| - |B \cap C| $ 
+
+$ B \cup C = |B| + |C| - |B \cap C| $
+
 $ B \cup C = \binom{17}{5} \cdot 2^{28} + \binom{28}{5} \cdot 2^{17} - \binom{17}{5} \cdot \binom{28}{5} $

src/content/questions/comp2804/2017-winter-final/4/solution.md

@@ -1,11 +1,21 @@
 Let A be the event that a string of length n is a palindrome
+
 The leftmost bit can be 1 or 0: 2 possibilities
+
 The rightmost bit must be the same as the leftmost bit: 1 possibility
+
 The second leftmost bit can be 1 or 0: 2 possibilities
+
 The second rightmost bit must be the same as the second leftmost bit: 1 possibility
+
 ...
+
 The innermost leftmost bit can be 1 or 0: 2 possibilities
+
 The innermost rightmost bit must be the same as the innermost leftmost bit: 1 possibility
+
 $ |A| = 2 \cdot 1 \cdot 2 \cdot 1 \cdot ... \cdot 2 \cdot 1 = 2^{n/2} $
+
 Usually, this would be right; however, we know the middle bit is always the same as itself AND n/2 is a \fraction since n is odd
+
 $ |A| = 2^{ \frac{n+1}{2}} $

src/content/questions/comp2804/2017-winter-final/6/solution.md

@@ -6,12 +6,21 @@
     <li> Let's determine $ \overline{B} \cap \overline{C} $ <br/> 
     $ | \overline{B} \cap \overline{C} | = 25 $
 </ul>
-Now, let's determine $ B \cup C $ 
-$ B \cup C = 100 - | \overline{B} \cap \overline{C} | $ 
-$ B \cup C = 100 - 25 $ 
-$ B \cup C = 75 $ 
-Finally, we can determine $ B \cap C $ 
-$ B \cup C = |B| + |C| - |B \cap C| $ 
-$ 75 = 63 + 71 - |B \cap C| $ 
-$ 75 = 134 - |B \cap C| $ 
+
+Now, let's determine $ B \cup C $
+
+$ B \cup C = 100 - | \overline{B} \cap \overline{C} | $
+
+$ B \cup C = 100 - 25 $
+
+$ B \cup C = 75 $
+
+Finally, we can determine $ B \cap C $
+
+$ B \cup C = |B| + |C| - |B \cap C| $
+
+$ 75 = 63 + 71 - |B \cap C| $
+
+$ 75 = 134 - |B \cap C| $
+
 $ |B \cap C| = 59 $

src/content/questions/comp2804/2017-winter-final/7/solution.md

@@ -1,4 +1,7 @@
 Let's just write down the possible bitstrings and add them up
-$ 0, 0, S*{n-2} $
-$ 1, 0, 0, 0, S*{n-4} $
+
+$ 0, 0, S\*{n-2} $
+
+$ 1, 0, 0, 0, S\*{n-4} $
+
 Adding them up, we get $ S*n = S*{n-2} + S\_{n-4} $

src/content/questions/comp2804/2017-winter-final/8/solution.md

@@ -1,7 +1,13 @@
 Consider the case where the current bit, next bit, and bit after that are all 0
+
 That means we can just calculate the remaining possible bitstring combinations since the requirements have already been met
+
 $ 2^{n-3} $
-$ 1, S*{n-1} $
-$ 0, 1, S*{n-2} $
-$ 0, 0, 1, S*{n-3} $
-Adding them up, we get $ S_n = S*{n-1} + S*{n-2} + S*{n-3} + 2^{n-3} $
+
+$ 1, S\*{n-1} $
+
+$ 0, 1, S\*{n-2} $
+
+$ 0, 0, 1, S\*{n-3} $
+
+Adding them up, we get $ S_n = S*{n-1} + S*{n-2} + S\*{n-3} + 2^{n-3} $