spaced out 2016 fall final

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 $

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} $

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} $

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} $

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>

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} $

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}} $

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} $

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} $

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!} $

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} $

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} $

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.

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.

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 $

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} $