diff --git a/src/content/questions/comp2804/2018-winter-final/1/solution.md b/src/content/questions/comp2804/2018-winter-final/1/solution.md index d213aff1..801aa4a5 100644 --- a/src/content/questions/comp2804/2018-winter-final/1/solution.md +++ b/src/content/questions/comp2804/2018-winter-final/1/solution.md @@ -1,3 +1,5 @@ Let's choose 12 beer bottles out of the 20: $ \binom{20}{12} $ + Let's choose 18 cider bottles out of the 50: $ \binom{50}{18} $ + $ \binom{20}{12} \cdot \binom{50}{18} $ diff --git a/src/content/questions/comp2804/2018-winter-final/10/solution.md b/src/content/questions/comp2804/2018-winter-final/10/solution.md index 4d9de464..9023aeae 100644 --- a/src/content/questions/comp2804/2018-winter-final/10/solution.md +++ b/src/content/questions/comp2804/2018-winter-final/10/solution.md @@ -1,8 +1,15 @@ -On SundayEveningExam$ (n) $, the line \enquote{I don't like Sunday evening exams} is printed n times -On SundayEveningExam$ (n-1) $, the line \enquote{I don't like Sunday evening exams} is printed n-1 times -On SundayEveningExam$ (n-2) $, the line \enquote{I don't like Sunday evening exams} is printed n-2 times -... -On SundayEveningExam$ (1) $, the line \enquote{I don't like Sunday evening exams} is printed 1 time +On SundayEveningExam$ (n) $, the line \enquote{I don't like Sunday evening exams} is printed n times + +On SundayEveningExam$ (n-1) $, the line \enquote{I don't like Sunday evening exams} is printed n-1 times + +On SundayEveningExam$ (n-2) $, the line \enquote{I don't like Sunday evening exams} is printed n-2 times + +... + +On SundayEveningExam$ (1) $, the line \enquote{I don't like Sunday evening exams} is printed 1 time + $ P(n) = n + (n-1) + (n-2) + \text{...} + 1$ + Using the 1st Formula given on the Exam Sheet, it translates the above to the following: + $ P(n) = \frac{n(n+1)}{2} $ diff --git a/src/content/questions/comp2804/2018-winter-final/11/solution.md b/src/content/questions/comp2804/2018-winter-final/11/solution.md index 88cd6d6e..bc078cff 100644 --- a/src/content/questions/comp2804/2018-winter-final/11/solution.md +++ b/src/content/questions/comp2804/2018-winter-final/11/solution.md @@ -1,6 +1,11 @@ First, we choose which 3 of the 18 rolls are 5: $ \binom{18}{3} $ -We also need to calculate the prbability of rolling a 5 exactly 3 times: ${( \frac{1}{6})}^3 $ -We also need to calculate the probability of not rolling a 5 exactly 15 times: ${( \frac{5}{6})}^{15} $ + +We also need to calculate the prbability of rolling a 5 exactly 3 times: ${( \frac{1}{6})}^3 $ + +We also need to calculate the probability of not rolling a 5 exactly 15 times: ${( \frac{5}{6})}^{15} $ + $ \binom{18}{3} \cdot {( \frac{1}{6})}^3 \cdot {( \frac{5}{6})}^{15} $ + $ = \binom{18}{3} \cdot ( \frac{1^3}{6^3}) \cdot ( \frac{5^{15}}{6^{15}}) $ + $ = \binom{18}{3} \cdot \frac{5^{15}}{6^{18}} $ diff --git a/src/content/questions/comp2804/2018-winter-final/12/solution.md b/src/content/questions/comp2804/2018-winter-final/12/solution.md index e59b8aeb..82e79f3d 100644 --- a/src/content/questions/comp2804/2018-winter-final/12/solution.md +++ b/src/content/questions/comp2804/2018-winter-final/12/solution.md @@ -1,4 +1,7 @@ To guarantee that a string is a palindrome in this case, we have to ensure that the first half of the string matches the second half (but reversed). + At each position in the "half of the string", there is a $\frac{1}{3}$ chance that the corresponding character in the other half of the string is the same since there are 3 character options (a, b, c) and only 1 of those can match. + Since the string is an odd integer, half of a string would be: $\frac{n-1}{2}$. So if $n = 3$, then half of the string would be $\frac{3-1}{2} = 1$. This makes sense, as we would check if the 1st and last character are the same, and the 2nd character we know would already be the same as itself. + This means that the total probability a string is a palindrome is: $( \frac{1}{3})^{n-1/2}$ diff --git a/src/content/questions/comp2804/2018-winter-final/13/solution.md b/src/content/questions/comp2804/2018-winter-final/13/solution.md index 7b5cf444..642bcb44 100644 --- a/src/content/questions/comp2804/2018-winter-final/13/solution.md +++ b/src/content/questions/comp2804/2018-winter-final/13/solution.md @@ -17,7 +17,11 @@ $ |A \cap B| = 15 $
$ Pr(A \cap B) = \frac{15}{32} $ -$ Pr(A|B) = \frac{Pr(A \cap B)}{Pr(B)} $ -$ Pr(A|B) = \frac{ \frac{15}{32}}{ \frac{1}{2}} $ -$ Pr(A|B) = \frac{15}{32} \cdot 2 $ + +$ Pr(A|B) = \frac{Pr(A \cap B)}{Pr(B)} $ + +$ Pr(A|B) = \frac{ \frac{15}{32}}{ \frac{1}{2}} $ + +$ Pr(A|B) = \frac{15}{32} \cdot 2 $ + $ Pr(A|B) = \frac{15}{16} $ diff --git a/src/content/questions/comp2804/2018-winter-final/14/solution.md b/src/content/questions/comp2804/2018-winter-final/14/solution.md index 3220cd77..e0928c3f 100644 --- a/src/content/questions/comp2804/2018-winter-final/14/solution.md +++ b/src/content/questions/comp2804/2018-winter-final/14/solution.md @@ -1,9 +1,17 @@ The difference in string $a-b$ is non-zero at each position only if $a_i$ is 1 and $b_i$ is 0 or $a_i$ is 0 and $b_i$ is 1, for all $1 leq i leq 77$ + Let $X_i$ be 1 if $a_i$ and $b_i$ are different and 0 otherwise + $ Pr(X_i = 1) = \frac{1}{4} \cdot \frac{3}{4} + \frac{3}{4} \cdot \frac{1}{4} $ + $ Pr(X_i = 1) = \frac{3}{16} + \frac{3}{16} $ + $ Pr(X_i = 1) = \frac{6}{16} $ + $ Pr(X_i = 1) = \frac{3}{8} $ + The probability that each element in the string is non-zero is the probability that each element is different + $ \frac{3}{8} \cdot \frac{3}{8} \cdot \text{...} \cdot \frac{3}{8} $ + $ = {( \frac{3}{8})}^{77} $ diff --git a/src/content/questions/comp2804/2018-winter-final/15/solution.md b/src/content/questions/comp2804/2018-winter-final/15/solution.md index cbf950b3..f9742822 100644 --- a/src/content/questions/comp2804/2018-winter-final/15/solution.md +++ b/src/content/questions/comp2804/2018-winter-final/15/solution.md @@ -20,8 +20,13 @@ We'll take a slow and systematic approach to this question $ |A \cap B| = 1 \cdot 2 = 2 $
$ Pr(A \cap B) = \frac{2}{16} = \frac{1}{8} $ -Now, let's check whether it's independent -$ Pr(A \cap B) = Pr(A) \cdot Pr(B) $ -$ \frac{1}{8} = \frac{1}{4} \cdot \frac{1}{4} $ -$ \frac{1}{8} = \frac{1}{16} $ + +Now, let's check whether it's independent + +$ Pr(A \cap B) = Pr(A) \cdot Pr(B) $ + +$ \frac{1}{8} = \frac{1}{4} \cdot \frac{1}{4} $ + +$ \frac{1}{8} = \frac{1}{16} $ + Since the two sides are not equal, the events are not independent diff --git a/src/content/questions/comp2804/2018-winter-final/16/solution.md b/src/content/questions/comp2804/2018-winter-final/16/solution.md index 56c1539f..a1f7f32d 100644 --- a/src/content/questions/comp2804/2018-winter-final/16/solution.md +++ b/src/content/questions/comp2804/2018-winter-final/16/solution.md @@ -22,8 +22,13 @@ We'll take a slow and systematic approach to this question $ |A \cap B| = 1 $
$ Pr(A \cap B) = \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}{8} \cdot \frac{13}{16} $ -$ \frac{1}{32} = \frac{13}{128} $ + +Now, let's check whether it's independent + +$ Pr(A \cap B) = Pr(A) \cdot Pr(B) $ + +$ \frac{1}{32} = \frac{1}{8} \cdot \frac{13}{16} $ + +$ \frac{1}{32} = \frac{13}{128} $ + Since the two sides are not equal, the events are not independent diff --git a/src/content/questions/comp2804/2018-winter-final/17/solution.md b/src/content/questions/comp2804/2018-winter-final/17/solution.md index d1cc250c..5ff9c0dc 100644 --- a/src/content/questions/comp2804/2018-winter-final/17/solution.md +++ b/src/content/questions/comp2804/2018-winter-final/17/solution.md @@ -1,4 +1,7 @@ Let S be the total number of ways to choose a permutation from a set of size $2n$: $2n!$ + For the first element and the last element in the permutation are even integers, we have to choose 2 even numbers. There are two ways to rearrange the 2 even numbers: $\binom{n}{2} \cdot 2$ + For the remaining positions in the permutation, we have $(2n-2)!$ ways to rearrange these numbers: $(2n-2)!$. + $Pr(A) = \frac{|A|}{|S|} = \frac{\binom{n}{2} \cdot 2 \cdot (2n-2)!}{2n!} = \frac{ \frac{n!}{2!(n-2)!} \cdot 2 \cdot (2n-2)!}{2n!} = \frac{n(n-1)}{2n(2n-1)} = \frac{n-1}{2(2n-1)}$ diff --git a/src/content/questions/comp2804/2018-winter-final/18/solution.md b/src/content/questions/comp2804/2018-winter-final/18/solution.md index 62dbfbaf..50006182 100644 --- a/src/content/questions/comp2804/2018-winter-final/18/solution.md +++ b/src/content/questions/comp2804/2018-winter-final/18/solution.md @@ -11,6 +11,9 @@ $ { (0, 0), (0, 1), (1, 0), (1, 1) } $
  • For $ (1,1) $, the minimum value is 1
    There is a $ \frac{1}{4} $ chance of this happening -$ \mathbb{E}(Z) = 0 \cdot \frac{1}{4} + 0 \cdot \frac{1}{4} + 0 \cdot \frac{1}{4} + 1 \cdot \frac{1}{4} $ -$ \mathbb{E}(Z) = 0 + 0 + 0 + \frac{1}{4} $ + +$ \mathbb{E}(Z) = 0 \cdot \frac{1}{4} + 0 \cdot \frac{1}{4} + 0 \cdot \frac{1}{4} + 1 \cdot \frac{1}{4} $ + +$ \mathbb{E}(Z) = 0 + 0 + 0 + \frac{1}{4} $ + $ \mathbb{E}(Z) = \frac{1}{4} $ diff --git a/src/content/questions/comp2804/2018-winter-final/19/solution.md b/src/content/questions/comp2804/2018-winter-final/19/solution.md index 5792fb22..3817cca4 100644 --- a/src/content/questions/comp2804/2018-winter-final/19/solution.md +++ b/src/content/questions/comp2804/2018-winter-final/19/solution.md @@ -1,10 +1,19 @@ Let $X_i$ be 1 if $b_i \cdot b_{i+1} = 0$ and 0 otherwise. This only happens if a pair of consecutive bits contain a 0. + $ Pr(X*i = 1) = Pr{(b_i = 0 \text{ and } b*{i+1} = 1)} + Pr{(b*i = 1 \text{ and } b*{i+1} = 0)} + Pr{(b*i = 0 \text{ and } b*{i+1} = 0)} $ -$ Pr(X*i = 1) = \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} $ + +$ Pr(X\*i = 1) = \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} $ + $ Pr(X_i = 1) = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} $ + $ Pr(X_i = 1) = \frac{3}{4} $ + Since the last bit has no bit after it, we only take into account the first $n-1$ bits + $ \mathbb{E}(X) = \sum*{k=1}^{n-1} 1 \cdot Pr(X*i = 1) $ -$ \mathbb{E}(X) = \sum*{k=1}^{n-1} \frac{3}{4} $ + +$ \mathbb{E}(X) = \sum\*{k=1}^{n-1} \frac{3}{4} $ + $ \mathbb{E}(X) = \frac{3}{4} \cdot (n-1) $ + $ \mathbb{E}(X) = \frac{3(n-1)}{4} $ diff --git a/src/content/questions/comp2804/2018-winter-final/2/solution.md b/src/content/questions/comp2804/2018-winter-final/2/solution.md index aecc216a..56bf66ff 100644 --- a/src/content/questions/comp2804/2018-winter-final/2/solution.md +++ b/src/content/questions/comp2804/2018-winter-final/2/solution.md @@ -13,6 +13,9 @@ First, choose 12 cider bottles out of the 50: $ \binom{50}{12} $
    $ |B \cap C| = \binom{20}{12} \cdot \binom{50}{12} $ -Now, we can find $ B \cup C $ -$ |B \cup C| = |B| + |C| - |B \cap C| $ + +Now, we can find $ B \cup C $ + +$ |B \cup C| = |B| + |C| - |B \cap C| $ + $ |B \cup C| = \binom{20}{12} \cdot 2^{50} + \binom{50}{12} \cdot 2^{20} - \binom{20}{12} \cdot \binom{50}{12} $ diff --git a/src/content/questions/comp2804/2018-winter-final/20/solution.md b/src/content/questions/comp2804/2018-winter-final/20/solution.md index a8c9ae51..5892f57e 100644 --- a/src/content/questions/comp2804/2018-winter-final/20/solution.md +++ b/src/content/questions/comp2804/2018-winter-final/20/solution.md @@ -1,9 +1,17 @@ Let $X_i$ be 1 if $C$ is not at position 1 and 2 if $C$ is at position 1 -$ Pr(X*i=1) = Pr(\text{$B_i$ is at position 1}) = \frac{1}{n+1} $ + +$ Pr(X\*i=1) = Pr(\text{$B_i$ is at position 1}) = \frac{1}{n+1} $ + $ Pr(X_i=2) = Pr(\text{$C$ is at position 1}) = \frac{1}{n+1} $ + $ \mathbb{E}(X) = \sum*{k=2}^{n+1} 1 \cdot Pr(X*i=1) + 2 \cdot Pr(X_i=2) $ -$ \mathbb{E}(X) = \sum*{k=2}^{n+1} 1 \cdot \frac{1}{n+1} + 2 \cdot \frac{1}{n+1} $ + +$ \mathbb{E}(X) = \sum\*{k=2}^{n+1} 1 \cdot \frac{1}{n+1} + 2 \cdot \frac{1}{n+1} $ + $ \mathbb{E}(X) = \sum\_{k=2}^{n+1} \frac{1}{n+1} + \frac{2}{n+1} $ + $ \mathbb{E}(X) = (n+1-2) \cdot ( \frac{1}{n+1} + \frac{2}{n+1})$ + $ \mathbb{E}(X) = \frac{n+1-2+1+2}{n+1} $ + $ \mathbb{E}(X) = \frac{n+2}{n+1} $ diff --git a/src/content/questions/comp2804/2018-winter-final/21/solution.md b/src/content/questions/comp2804/2018-winter-final/21/solution.md index aa55ee03..105ea739 100644 --- a/src/content/questions/comp2804/2018-winter-final/21/solution.md +++ b/src/content/questions/comp2804/2018-winter-final/21/solution.md @@ -1,5 +1,7 @@ For $X$ and $Z$ to be independent, $Pr(X = x \cap Z = z) = Pr(X = x) \cdot Pr(Z = z)$, for all real numbers $x$ and $z$, by definition of independent random variables. + If we can find a counterexample, aka where $Pr(X = x \cap Z = z) neq Pr(X = x) \cdot Pr(Z = z)$, then we know the two are not independent. + Let $z = 0$, $x = 0$ and $y = 0$: + Now we check if LHS = RHS for the expression: + + Since LHS is not equal to RHS, then the events $X$ and $Z$ are not independent. diff --git a/src/content/questions/comp2804/2018-winter-final/22/solution.md b/src/content/questions/comp2804/2018-winter-final/22/solution.md index 217b0db6..986e612c 100644 --- a/src/content/questions/comp2804/2018-winter-final/22/solution.md +++ b/src/content/questions/comp2804/2018-winter-final/22/solution.md @@ -1,4 +1,7 @@ In the algorithm, the value of $k$ keeps increasing if the two dice rolls are different. Theoreticaly, this algorithm can go on forever, until the two dice rolls are the same. + This is an example of the Geometric Distribution. + Let $p$ represent the probability of not ending the recursive algorithm. Using the formula for the expected value for Geometric Distribution we get: + $E(X) = \frac{1}{p} = \frac{1}{ \frac{5}{6}} = \frac{6}{5}$ diff --git a/src/content/questions/comp2804/2018-winter-final/23/solution.md b/src/content/questions/comp2804/2018-winter-final/23/solution.md index 567a98bc..d60b528a 100644 --- a/src/content/questions/comp2804/2018-winter-final/23/solution.md +++ b/src/content/questions/comp2804/2018-winter-final/23/solution.md @@ -1,7 +1,11 @@ To prove that this statement is false, we need to provide a counterexample: + Let X, Y, Z all represent the random variables where $X, Y, Z = 1$ if a coin toss is heads and $X, Y, Z = 0$ if a coin toss is tails. + The expected value of $X, Y, Z$ in this case (as\suming that $\frac{1}{2}$ chance of hitting heads or tails) is: $E(X) = E(Y) = E(Z) = 0 \cdot \frac{1}{2} + 1 \cdot \frac{1}{2} = \frac{1}{2}$. + This means that the $min(E(X), E(Y), E(Z) = min( \frac{1}{2}, \frac{1}{2}, \frac{1}{2}) = \frac{1}{2}$ is $\frac{1}{2}$. + For the LHS, we need to calculate the expected value for $min(X, Y, Z)$. There are only two possible values in this case: $min(X, Y, Z) = 0$ and $min(X, Y, Z) = 1$ + Since $E(min(X, Y, Z)) = \frac{1}{8}$ and $min(E(X), E(Y), E(Z)) = \frac{1}{2}$, the two are not the same, so the statement is false. diff --git a/src/content/questions/comp2804/2018-winter-final/24/solution.md b/src/content/questions/comp2804/2018-winter-final/24/solution.md index 5b1d4858..57ab19d9 100644 --- a/src/content/questions/comp2804/2018-winter-final/24/solution.md +++ b/src/content/questions/comp2804/2018-winter-final/24/solution.md @@ -1,5 +1,7 @@ Let S be the set of all possibilities = $ |S| = 8^{3} $ + Let A be the set of all possibilities where a student gets exactly 2 ciders + To get exactly 2 ciders, a student has to have 2 ciders and 1 beer. There are 3 ways to get 2 ciders and a beer: $CBC, BCC, CCB$. -Now, we can sum -$ Pr(A) = Pr( CBC ) + Pr( BCC ) + Pr( CCB ) $ -$ Pr(A) = (5^2 \cdot 3) + (5^2 \cdot 3) + (5^2 \cdot 3) $ -$ Pr(A) = 3 \cdot \frac{5^2 \cdot 3}{8^3} $ -$ Pr(A) = \frac{3^2 \cdot 5^2}{8^3}$ + +Now, we can sum + +$ Pr(A) = Pr( CBC ) + Pr( BCC ) + Pr( CCB ) $ + +$ Pr(A) = (5^2 \cdot 3) + (5^2 \cdot 3) + (5^2 \cdot 3) $ + +$ Pr(A) = 3 \cdot \frac{5^2 \cdot 3}{8^3} $ + +$ Pr(A) = \frac{3^2 \cdot 5^2}{8^3}$ + Let $X_i$ be an indicator random variable where: + $ 1 \text{ if a student gets exactly 2 ciders} $ + $ 0 \text{ otherwise} $ -$ Pr(X*i = 1) = Pr(A) = \frac{3^2 \cdot 5^2}{8^3}$ -$ \mathbb{E}(X) = \sum*{k=1}^{16} 1 \cdot Pr(X*i = 1) $ -$ \mathbb{E}(X) = \sum*{k=1}^{16} \frac{3^2 \cdot 5^2}{8^3} $ -$ \mathbb{E}(X) = 16 \cdot \frac{3^2 \cdot 5^2}{8^3} $ -$ \mathbb{E}(X) = 2^4 \cdot \frac{3^2 \cdot 5^2}{8^3} $ -$ \mathbb{E}(X) = \frac{2^4 \cdot 3^2 \cdot 5^2}{8^3} $ + +$ Pr(X\*i = 1) = Pr(A) = \frac{3^2 \cdot 5^2}{8^3}$ + +$ \mathbb{E}(X) = \sum*{k=1}^{16} 1 \cdot Pr(X*i = 1) $ + +$ \mathbb{E}(X) = \sum\*{k=1}^{16} \frac{3^2 \cdot 5^2}{8^3} $ + +$ \mathbb{E}(X) = 16 \cdot \frac{3^2 \cdot 5^2}{8^3} $ + +$ \mathbb{E}(X) = 2^4 \cdot \frac{3^2 \cdot 5^2}{8^3} $ + +$ \mathbb{E}(X) = \frac{2^4 \cdot 3^2 \cdot 5^2}{8^3} $ + $ \mathbb{E}(X) = 2^4 \cdot 3^2 \cdot \frac{5^2}{8^3} $ diff --git a/src/content/questions/comp2804/2018-winter-final/3/solution.md b/src/content/questions/comp2804/2018-winter-final/3/solution.md index 9abdd09c..bfbcea1c 100644 --- a/src/content/questions/comp2804/2018-winter-final/3/solution.md +++ b/src/content/questions/comp2804/2018-winter-final/3/solution.md @@ -12,8 +12,13 @@ We need to take into account all subsets of cider bottles that we add to the 2 beer bottles: $ 2^{50} $
    $ |C| = \binom{20}{2} \cdot 2^{50} $ -Since the question is asking for all possibilities excluding 0, 1, and 2 beer bottles, we can find the total number of possibilities by subtracting the number of possibilities that contain 0, 1, and 2 beer bottles from the total number of possibilities -Let D be the set of all bottles that contain at least 3 beer bottles -$ |D| = 2^{70} - |A| - |B| - |C| $ -$ |D| = 2^{70} - \binom{20}{0} \cdot 2^{50} - \binom{20}{1} \cdot 2^{50} - \binom{20}{2} \cdot 2^{50} $ + +Since the question is asking for all possibilities excluding 0, 1, and 2 beer bottles, we can find the total number of possibilities by subtracting the number of possibilities that contain 0, 1, and 2 beer bottles from the total number of possibilities + +Let D be the set of all bottles that contain at least 3 beer bottles + +$ |D| = 2^{70} - |A| - |B| - |C| $ + +$ |D| = 2^{70} - \binom{20}{0} \cdot 2^{50} - \binom{20}{1} \cdot 2^{50} - \binom{20}{2} \cdot 2^{50} $ + $ |D| = 2^{70} - 2^{50} - 20 \cdot 2^{50} - \binom{20}{2} \cdot 2^{50} $ diff --git a/src/content/questions/comp2804/2018-winter-final/4/solution.md b/src/content/questions/comp2804/2018-winter-final/4/solution.md index 6b99912b..d82f7641 100644 --- a/src/content/questions/comp2804/2018-winter-final/4/solution.md +++ b/src/content/questions/comp2804/2018-winter-final/4/solution.md @@ -12,7 +12,11 @@ We need to choose 5 positions out of the remaining 35 for the $c$'s: $ \binom{35}{5} $
    Then for the remaining characters in the strings, they can be either $b$ or $d$ = $ 2^{30} $ -Now, we can find $ A \cup B $ -$ |A \cup B| = |A| + |B| - |A \cap B| $ -$ |A \cup B| = \binom{40}{5} \cdot 3^{35} + \binom{40}{5} \cdot 3^{35} - \binom{40}{5} \cdot \binom{35}{5} \cdot 2^{30} $ + +Now, we can find $ A \cup B $ + +$ |A \cup B| = |A| + |B| - |A \cap B| $ + +$ |A \cup B| = \binom{40}{5} \cdot 3^{35} + \binom{40}{5} \cdot 3^{35} - \binom{40}{5} \cdot \binom{35}{5} \cdot 2^{30} $ + $ |A \cup B| = 2 \cdot \binom{40}{5} \cdot 3^{35} - \binom{40}{5} \cdot \binom{35}{5} \cdot 2^{30} $ diff --git a/src/content/questions/comp2804/2018-winter-final/5/solution.md b/src/content/questions/comp2804/2018-winter-final/5/solution.md index 69bddb5d..d948cfb8 100644 --- a/src/content/questions/comp2804/2018-winter-final/5/solution.md +++ b/src/content/questions/comp2804/2018-winter-final/5/solution.md @@ -1,4 +1,5 @@ The answer is C, and I will show how the expression in the question shows the number of ways to choose a 2-element subset from a set consisting of m + n elements: + From a set of $m + n$ elements, we can form 2-element subsets in 3 different ways: + Now, if we apply the Sum Rule to all of these cases above, we get $\binom{m}{2} + \binom{n}{2} + m \cdot n$, which matches the expression above. diff --git a/src/content/questions/comp2804/2018-winter-final/6/solution.md b/src/content/questions/comp2804/2018-winter-final/6/solution.md index ccd94580..73155cde 100644 --- a/src/content/questions/comp2804/2018-winter-final/6/solution.md +++ b/src/content/questions/comp2804/2018-winter-final/6/solution.md @@ -1,6 +1,9 @@ We can rephrase this into something more familiar. 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. + $ x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 = 25 $ + This is one of the divider questions + Assume we have 25 bananas and 6 dividers. We place the dividers between any set of bananas to indicate how many are eaten between dividers -Since we place dividers down, we need to add 6 dividers to the 25 bananas + +Since we place dividers down, we need to add 6 dividers to the 25 bananas + We choose 6 positions out of the 31 for the dividers: $ \binom{31}{6} $ diff --git a/src/content/questions/comp2804/2018-winter-final/7/solution.md b/src/content/questions/comp2804/2018-winter-final/7/solution.md index d53cf3f1..9f0bbaa9 100644 --- a/src/content/questions/comp2804/2018-winter-final/7/solution.md +++ b/src/content/questions/comp2804/2018-winter-final/7/solution.md @@ -15,7 +15,11 @@ Then we choose 1 of the remaining 1 positions to place the $OO$: $ \binom{1}{1} = 1 $
    $ |\overline{A}| = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 5! $ -Now, we can find the number of awesome strings -$|A| = |S| - |\overline{A}| $ -$|A| = (6 \cdot \binom{5}{2} \cdot 3 \cdot 2 \cdot 1) - 5!$ + +Now, we can find the number of awesome strings + +$|A| = |S| - |\overline{A}| $ + +$|A| = (6 \cdot \binom{5}{2} \cdot 3 \cdot 2 \cdot 1) - 5!$ + $|A| = (6 \cdot \binom{5}{2} \cdot 3 \cdot 2) - 5!$ diff --git a/src/content/questions/comp2804/2018-winter-final/8/solution.md b/src/content/questions/comp2804/2018-winter-final/8/solution.md index b812e661..4ecf977e 100644 --- a/src/content/questions/comp2804/2018-winter-final/8/solution.md +++ b/src/content/questions/comp2804/2018-winter-final/8/solution.md @@ -1,6 +1,11 @@ The first character can be any of the 4 characters: 4 + The second character can be any of the 3 characters that are not the first character: 3 + The third character can be any of the 3 characters that are not the second character: 3 + ... + The nth character can be any of the 3 characters that are not the $(n-1)$th character: 3 + $= 4 \cdot 3^{n-1} $ diff --git a/src/content/questions/comp2804/2018-winter-final/9/solution.md b/src/content/questions/comp2804/2018-winter-final/9/solution.md index 4a81b490..4af1c276 100644 --- a/src/content/questions/comp2804/2018-winter-final/9/solution.md +++ b/src/content/questions/comp2804/2018-winter-final/9/solution.md @@ -5,5 +5,7 @@ Let's write out the possibilities and sum them:
  • $ 0, S*{n-1} $
  • $ 1, 0, S_{n-2} $ -Although this takes into account many possibilities, it does not take into a bitstring that is made of all 1's: 1 -$ S*n = S*{n-1} + S_{n-2} + 1 $ + +Although this takes into account many possibilities, it does not take into a bitstring that is made of all 1's: 1 + +$ S*n = S*{n-1} + S\_{n-2} + 1 $