added more solutions to 2014 fall final

src/content/questions/comp2804/2014-fall-final/10/solution.md

@@ -1,24 +1,28 @@
 Let's calculate $f(1)$ first
-$f(1) = f(0) + 8 \cdot 1 - 2 $ 
+
+$f(1) = f(0) + 8 \cdot 1 - 2 $
+
 $f(1) = -17 + 8 - 2 $
+
 $f(1) = -11 $
 
 <ul>
-<li> $ f(n) = 3n^{2} - n - 17 $ 
-$ f(1) = 3 \cdot 1^2 - 1 - 17 $ 
-$ f(1) = 3 - 1 - 17 $ 
-$ f(1) = -15 $
-<li> $ f(n) = 3n^{2} + n - 17 $ 
-$ f(1) = 3 \cdot 1^2 + 1 - 17 $ 
-$ f(1) = 3 + 1 - 17 $ 
-$ f(1) = -13 $
-<li> $ f(n) = 4n^{2} - 2n - 17 $ 
-$ f(1) = 4 \cdot 1^2 - 2 - 17 $ 
-$ f(1) = 4 - 2 - 17 $ 
-$ f(1) = -15 $
-<li> $ f(n) = 4n^{2} + 2n - 17 $ 
-$ f(1) = 4 \cdot 1^2 + 2 - 17 $ 
-$ f(1) = 4 + 2 - 17 $ 
-$ f(1) = -11 $
+    <li> $ f(n) = 3n^{2} - n - 17 $ <br/>
+    $ f(1) = 3 \cdot 1^2 - 1 - 17 $ <br/>
+    $ f(1) = 3 - 1 - 17 $ <br/>
+    $ f(1) = -15 $
+    <li> $ f(n) = 3n^{2} + n - 17 $ <br/>
+    $ f(1) = 3 \cdot 1^2 + 1 - 17 $ <br/>
+    $ f(1) = 3 + 1 - 17 $ <br/>
+    $ f(1) = -13 $
+    <li> $ f(n) = 4n^{2} - 2n - 17 $ <br/>
+    $ f(1) = 4 \cdot 1^2 - 2 - 17 $ <br/>
+    $ f(1) = 4 - 2 - 17 $ <br/>
+    $ f(1) = -15 $
+    <li> $ f(n) = 4n^{2} + 2n - 17 $ <br/>
+    $ f(1) = 4 \cdot 1^2 + 2 - 17 $ <br/>
+    $ f(1) = 4 + 2 - 17 $ <br/>
+    $ f(1) = -11 $
 </ul>
+
 The correct answer is $ f(n) = 4n^{2} + 2n - 17 $ because it is the only one that gives the correct value for $ f(1) $

src/content/questions/comp2804/2014-fall-final/12/solution.md

@@ -1,17 +1,20 @@
 <ul>
-<li> Let A = Event that both balls are blue 
-A occurs when we choose 2 of the 5 blue balls 
-$ |A| = \binom{5}{2} $
-	<li> Let B = Event that both balls are red 
-	      B occurs when we choose 2 of the 4 red balls 
-	      $ |B| = \binom{4}{2} $
-	<li> Let C = Event that both balls are green 
-	      C occurs when we choose 2 of the 6 green balls 
-	      $ |C| = \binom{6}{2} $
-	<li> Let S = All possible outcomes 
-	      S occurs when we choose 2 of the 15 balls 
-	      $ |S| = \binom{15}{2} $
+	<li> Let A = Event that both balls are blue <br/>
+	A occurs when we choose 2 of the 5 blue balls <br/>
+	$ |A| = \binom{5}{2} $
+	<li> Let B = Event that both balls are red <br/>
+	B occurs when we choose 2 of the 4 red balls <br/>
+	$ |B| = \binom{4}{2} $
+	<li> Let C = Event that both balls are green <br/>
+	C occurs when we choose 2 of the 6 green balls <br/>
+	$ |C| = \binom{6}{2} $
+	<li> Let S = All possible outcomes <br/>
+	S occurs when we choose 2 of the 15 balls <br/>
+	$ |S| = \binom{15}{2} $
 </ul>
-$ |A \cup B \cup C| = |A| + |B| + |C| $ 
-$ |A \cup B \cup C| = \binom{5}{2} + \binom{4}{2} + \binom{6}{2} $ 
+
+$ |A \cup B \cup C| = |A| + |B| + |C| $
+
+$ |A \cup B \cup C| = \binom{5}{2} + \binom{4}{2} + \binom{6}{2} $
+
 $ Pr(A \cup B \cup C) = \frac{\binom{5}{2} + \binom{4}{2} + \binom{6}{2}}{\binom{15}{2}} $

src/content/questions/comp2804/2014-fall-final/13/solution.md

@@ -1,9 +1,17 @@
 Let A = Event that Annie, Boris, and Charlie have the same birthday
+
 Annie has a birthday on any of the 365 days: 365
+
 Boris would only then have a birthday on the same day as Annie: 1
+
 Charlie would only then have a birthday on the same day as Annie and Boris: 1
+
 $ |A| = 365 \cdot 1 \cdot 1 $
+
 Let S = All possible outcomes
+
 $ |S| = 365^3 $
+
 $ Pr(A) = \frac{365}{365^3} $
+
 $ Pr(A) = \frac{1}{365^2} $

src/content/questions/comp2804/2014-fall-final/14/solution.md

@@ -1,12 +1,12 @@
 Let's go through each option and explain
 
 <ul>
-<li> $1 - (1/2)^{7}$ 
-This is just the complement rule, which usually works but they're subtracting the probability that we get 6 tails followed by a head 
-It might've worked if they were subtracting the probability that we get 7 tails followed by a head from the whole set of possibilities
-<li> $\sum_{k=0}^{7} (1/2)^{k}$ 
-This one is wrong because it starts at $ { left( \frac{1}{2} right)}^0 $ but it doesn't take into account the fact that we have to flip a head to stop, which has a probability of $ \frac{1}{2} $
-	<li> $\sum_{k=0}^{7} (1/2)^{k+1}$ 
-This one is beautiful 
-It says, probability of flipping a heads right away, probability of flipping a tails folloewd by a heads, probability of flipping 2 tails followed by a heads, and so on until 7 tails followed by a heads
+	<li> $1 - (1/2)^{7}$ <br/>
+	This is just the complement rule, which usually works but they're subtracting the probability that we get 6 tails followed by a head <br/>
+	It might've worked if they were subtracting the probability that we get 7 tails followed by a head from the whole set of possibilities
+	<li> $\sum_{k=0}^{7} (1/2)^{k}$ <br/>
+	This one is wrong because it starts at $ { left( \frac{1}{2} right)}^0 $ but it doesn't take into account the fact that we have to flip a head to stop, which has a probability of $ \frac{1}{2} $
+	<li> $\sum_{k=0}^{7} (1/2)^{k+1}$ <br/>
+	This one is beautiful <br/>
+	It says, probability of flipping a heads right away, probability of flipping a tails folloewd by a heads, probability of flipping 2 tails followed by a heads, and so on until 7 tails followed by a heads
 </ul>

src/content/questions/comp2804/2014-fall-final/15/solution.md

@@ -1,12 +1,14 @@
 <ul>
-<li> Let A = Event that both balls are red 
-A occurs when we choose 2 of the 4 red balls 
-$ |A| = \binom{4}{2} $
-	<li> Let B = Event that both balls have the same color 
-	      B occurs when we choose 2 of the 5 blue balls or 2 of the 4 red balls 
-	      $ |B| = \binom{5}{2} + \binom{4}{2} $
-	<li> Let $A \cap B$ = Event that both balls are red and both balls have the same color 
-$ |A \cap B| = \binom{4}{2} $
+	<li> Let A = Event that both balls are red <br/>
+	A occurs when we choose 2 of the 4 red balls <br/>
+	$ |A| = \binom{4}{2} $
+	<li> Let B = Event that both balls have the same color <br/>
+	B occurs when we choose 2 of the 5 blue balls or 2 of the 4 red balls <br/>
+	$ |B| = \binom{5}{2} + \binom{4}{2} $
+	<li> Let $A \cap B$ = Event that both balls are red and both balls have the same color <br/>
+	$ |A \cap B| = \binom{4}{2} $
 </ul>
-$ Pr(A|B) = \frac{|A \cap B|}{|B|} $ 
+
+$ Pr(A|B) = \frac{|A \cap B|}{|B|} $
+
 $ Pr(A|B) = \frac{\binom{4}{2}}{\binom{5}{2} + \binom{4}{2}} $

src/content/questions/comp2804/2014-fall-final/16/solution.md

@@ -1,12 +1,14 @@
 <ul>
-<li> Let A = Event that $x$ is even 
-A occurs when we choose an even number from the set ${2, 4, 6, 8, 10}$ 
-$ |A| = 5 $
-	<li> Let B = Event that $x$ is divisible by 3 
-B occurs when we choose a number from the set ${3, 6, 9}$ 
-$ |B| = 3 $
-	<li> Let $A \cap B$ = Event that $x$ is even and divisible by 3 
-$ |A \cap B| = 1 $
+	<li> Let A = Event that $x$ is even <br/>
+	A occurs when we choose an even number from the set ${2, 4, 6, 8, 10}$ <br/>
+	$ |A| = 5 $
+	<li> Let B = Event that $x$ is divisible by 3 <br/>
+	B occurs when we choose a number from the set ${3, 6, 9}$ <br/>
+	$ |B| = 3 $
+	<li> Let $A \cap B$ = Event that $x$ is even and divisible by 3 <br/>
+	$ |A \cap B| = 1 $
 </ul>
-$ Pr(A|B) = \frac{|A \cap B|}{|B|} $ 
+
+$ Pr(A|B) = \frac{|A \cap B|}{|B|} $
+
 $ Pr(A|B) = \frac{1}{3} $

src/content/questions/comp2804/2014-fall-final/17/solution.md

@@ -1,17 +1,21 @@
 <ul>
-<li> Let S = All possible outcomes 
-$ |S| = 10 $
-	<li> A occurs when we choose an even number from the set ${2, 4, 6, 8, 10}$ 
-$ |A| = 5 $ 
-$Pr(A) = \frac{5}{10} = \frac{1}{2} $
-	<li> B occurs when we choose a number from the set ${1, 2, 3, 4, 5}$ 
-$ |B| = 5 $ 
-$Pr(B) = \frac{5}{10} = \frac{1}{2} $
-	<li> $A \cap B$ occurs when we choose an even number from the set ${2, 4}$ 
-$ |A \cap B| = 2 $ 
-$Pr(A \cap B) = \frac{2}{10} = \frac{1}{5} $
+	<li> Let S = All possible outcomes <br/>
+	$ |S| = 10 $
+	<li> A occurs when we choose an even number from the set ${2, 4, 6, 8, 10}$ <br/>
+	$ |A| = 5 $ <br/>
+	$Pr(A) = \frac{5}{10} = \frac{1}{2} $
+	<li> B occurs when we choose a number from the set ${1, 2, 3, 4, 5}$ <br/>
+	$ |B| = 5 $ <br/>
+	$Pr(B) = \frac{5}{10} = \frac{1}{2} $
+	<li> $A \cap B$ occurs when we choose an even number from the set ${2, 4}$ <br/>
+	$ |A \cap B| = 2 $ <br/>
+	$Pr(A \cap B) = \frac{2}{10} = \frac{1}{5} $
 </ul>
-$ 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} $ 
+
+$ 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} $
+
 Because the equation is false, these events are not independent.

src/content/questions/comp2804/2014-fall-final/18/solution.md

@@ -1,17 +1,21 @@
 <ul>
-<li> Let S = All possible outcomes 
-$ |S| = 10 $
-	<li> A occurs when we choose an even number from the set ${2, 4, 6, 8, 10}$ 
-$ |A| = 5 $ 
-$Pr(A) = \frac{5}{10} = \frac{1}{2} $
-	<li> B occurs when we choose a number from the set ${1, 2, 3, 4, 5, 6}$ 
-$ |B| = 6 $ 
-$Pr(B) = \frac{6}{10} = \frac{3}{5} $
-	<li> $A \cap B$ occurs when we choose an even number from the set ${2, 4, 6}$ 
-$ |A \cap B| = 3 $ 
-$Pr(A \cap B) = \frac{3}{10} $
+	<li> Let S = All possible outcomes <br/>
+	$ |S| = 10 $
+	<li> A occurs when we choose an even number from the set ${2, 4, 6, 8, 10}$ <br/>
+	$ |A| = 5 $ <br/>
+	$Pr(A) = \frac{5}{10} = \frac{1}{2} $
+	<li> B occurs when we choose a number from the set ${1, 2, 3, 4, 5, 6}$ <br/>
+	$ |B| = 6 $ <br/>
+	$Pr(B) = \frac{6}{10} = \frac{3}{5} $
+	<li> $A \cap B$ occurs when we choose an even number from the set ${2, 4, 6}$ <br/>
+	$ |A \cap B| = 3 $ <br/>
+	$Pr(A \cap B) = \frac{3}{10} $
 </ul>
-$ 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} $ 
+
+$ 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} $
+
 Because the equation is true, these events are independent.

src/content/questions/comp2804/2014-fall-final/19/solution.md

@@ -1,17 +1,21 @@
 We need to determine if events $ A $ and $ B $ are independent.
 
 <ul>
-<li> $ Pr(A) = \frac{1}{2} $
-	<li> If they have at least 3 kids and the third child is a boy, let's calculate what's needed to get there 
-	      Boy, Girl, Boy: $ \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8} $ 
-	      Girl, Boy, Boy: $ \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8} $ 
-	      $ Pr(B) = \frac{1}{8} + \frac{1}{8} = \frac{1}{4} $
-	<li> $ Pr(A \cap B) $ now 
-	      If the second child is a boy and we want them to keep pumping out babies, then the first two children must be of different genders 
-	      That leaves us with Girl, Boy, Boy: $ \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8} $ 
-	      $ Pr(A \cap B) = \frac{1}{8} $
+	<li> $ Pr(A) = \frac{1}{2} $
+	<li> If they have at least 3 kids and the third child is a boy, let's calculate what's needed to get there <br/>
+	Boy, Girl, Boy: $ \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8} $ <br/>
+	Girl, Boy, Boy: $ \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8} $ <br/>
+	$ Pr(B) = \frac{1}{8} + \frac{1}{8} = \frac{1}{4} $
+	<li> $ Pr(A \cap B) $ now <br/>
+	If the second child is a boy and we want them to keep pumping out babies, then the first two children must be of different genders <br/>
+	That leaves us with Girl, Boy, Boy: $ \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8} $ <br/>
+	$ Pr(A \cap B) = \frac{1}{8} $
 </ul>
-Now, let's check if $ Pr(A) \cdot Pr(B) = Pr(A \cap B) $ 
-$ \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8} $ 
-$ \frac{1}{8} = \frac{1}{8} $ 
+
+Now, let's check if $ Pr(A) \cdot Pr(B) = Pr(A \cap B) $
+
+$ \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8} $
+
+$ \frac{1}{8} = \frac{1}{8} $
+
 The events $ A $ and $ B $ are independent. Thus, the statement is true.

src/content/questions/comp2804/2014-fall-final/2/solution.md

@@ -1,17 +1,19 @@
 <ul>
-<li> A = Event that the password contains 0 event digits 
-Since every digit is odd, each of the 13 digits must be chosen from the set {1,3,5,7,9} 
-$ |A| = 5^{13} $
-	<li> B = Event that the password contains 1 event digit 
-	      Since 1 of the digits is even, there are 13 positions to place the single even digit: $ \binom{13}{1} $ 
-	      Since the 1 even digit can be chosen from the set {0,2,4,6,8}, there are 5 choices for the even digit: $ 5 $ 
-	      The remaining 12 digits must be chosen from the set {1,3,5,7,9}, so there are $ 5^{12} $ ways to choose the remaining digits 
-	      $ |B| = \binom{13}{1} \cdot 5 \cdot 5^{12} $
-	<li>  C = Event that the password contains 2 event digits 
-	      Since 2 of the digits are even, there are $ \binom{13}{2} $ ways to choose the 2 positions for the even digits 
-	      Since the 2 even digits can be chosen from the set {0,2,4,6,8}, each digit has 5 choices: $ 5^2 $ 
-	      The remaining 11 digits must be chosen from the set {1,3,5,7,9}, so there are $ 5^{11} $ ways to choose the remaining digits 
-	      $ |C| = \binom{13}{2} \cdot 5^2 \cdot 5^{11} $
+	<li> A = Event that the password contains 0 event digits <br/>
+	Since every digit is odd, each of the 13 digits must be chosen from the set {1,3,5,7,9} <br/>
+	$ |A| = 5^{13} $
+	<li> B = Event that the password contains 1 event digit <br/>
+	Since 1 of the digits is even, there are 13 positions to place the single even digit: $ \binom{13}{1} $ <br/>
+	Since the 1 even digit can be chosen from the set {0,2,4,6,8}, there are 5 choices for the even digit: $ 5 $ <br/>
+	The remaining 12 digits must be chosen from the set {1,3,5,7,9}, so there are $ 5^{12} $ ways to choose the remaining digits<br/> 
+	$ |B| = \binom{13}{1} \cdot 5 \cdot 5^{12} $
+	<li>  C = Event that the password contains 2 event digits <br/>
+	Since 2 of the digits are even, there are $ \binom{13}{2} $ ways to choose the 2 positions for the even digits <br/>
+	Since the 2 even digits can be chosen from the set {0,2,4,6,8}, each digit has 5 choices: $ 5^2 $ <br/>
+	The remaining 11 digits must be chosen from the set {1,3,5,7,9}, so there are $ 5^{11} $ ways to choose the remaining digits <br/>
+	$ |C| = \binom{13}{2} \cdot 5^2 \cdot 5^{11} $
 </ul>
-$ |A| + |B| + |C| = 5^{13} + \binom{13}{1} \cdot 5 \cdot 5^{12} + \binom{13}{2} \cdot 5^2 \cdot 5^{11} $ 
+
+$ |A| + |B| + |C| = 5^{13} + \binom{13}{1} \cdot 5 \cdot 5^{12} + \binom{13}{2} \cdot 5^2 \cdot 5^{11} $
+
 $ |A| + |B| + |C| = 5^{13} + 13 \cdot 5^{13} + \binom{13}{2} \cdot 5^{13} $

src/content/questions/comp2804/2014-fall-final/20/solution.md

@@ -0,0 +1,3 @@
+I mean, there are a total of 4 options, and only one of them is right.
+
+The probability of getting the right answer is $ \frac{1}{4} $.

src/content/questions/comp2804/2014-fall-final/21/solution.md

@@ -1,48 +1,43 @@
-subsection\*{Step-by-Step Solution}
-
-1. **Possible Outcomes and Corresponding Values:**
+1. **Possible Outcomes and Corresponding Values:**<br/>
    There are four possible outcomes when flipping two fair coins:
-   begin{itemize}
-   <li> First coin is heads (H), second coin is heads (H): $(H, H)$
-   <li> First coin is heads (H), second coin is tails (T): $(H, T)$
-   <li> First coin is tails (T), second coin is heads (H): $(T, H)$
-   <li> First coin is tails (T), second coin is tails (T): $(T, T)$
-   end{itemize}
+   <ul>
+      <li> First coin is heads (H), second coin is heads (H): $(H, H)$
+      <li> First coin is heads (H), second coin is tails (T): $(H, T)$
+      <li> First coin is tails (T), second coin is heads (H): $(T, H)$
+      <li> First coin is tails (T), second coin is tails (T): $(T, T)$
+   </ul>
 
-2. **Winning Amount for Each Outcome:**
+2. **Winning Amount for Each Outcome:**<br/>
    For each outcome, we determine the amount $ X $ that you win:
-   begin{itemize}
-   <li> $(H, H)$: The first coin is heads, so we don't lose. The second coin is heads, so we win 1 dollar. Thus, $ X = 1 $.
-   <li> $(H, T)$: The first coin is heads, so we don't lose. The second coin is tails, so we don't win. Thus, $ X = 0 $.
-   <li> $(T, H)$: The first coin is tails, so we lose (i.e., we win $-1$ dollars). The second coin is heads, so we win 1 dollar, but since the first coin being tails means we lose, this overrides the second coin's result. Thus, $ X = -1 $.
-   <li> $(T, T)$: The first coin is tails, so we lose (i.e., we win $-1$ dollars). The second coin is tails, so we don't win. Thus, $ X = -1 $.
-   end{itemize}
+   <ul>
+      <li> $(H, H)$: The first coin is heads, so we don't lose. <br/>
+      The second coin is heads, so we win 1 dollar. <br/>
+      Thus, $ X = 1 $.
+      <li> $(H, T)$: The first coin is heads, so we don't lose. <br/>
+      The second coin is tails, so we don't win. <br/>
+      Thus, $ X = 0 $.
+      <li> $(T, H)$: The first coin is tails, so we lose (i.e., we win $-1$ dollars). <br/>
+      The second coin is heads, so we win 1 dollar, but since the first coin being tails means we lose, this overrides the second coin's result. <br/>
+      Thus, $ X = -1 $.
+      <li> $(T, T)$: The first coin is tails, so we lose (i.e., we win $-1$ dollars). <br/>
+      The second coin is tails, so we don't win. Thus, $ X = -1 $.
+   </ul>
 
-3. **Calculate the Expected Value $ E(X) $:**
-The expected value $ E(X) $ is calculated using the formula:
-[
-E(X) = sum_{i} P(X = x_i) \cdot x_i
-]
-Each of the four outcomes has a probability of $ \frac{1}{4} $, as the coins are fair and independent.
+3. **Calculate the Expected Value $ E(X) $:**<br/>
+The expected value $ E(X) $ is calculated using the formula:<br/>
+$E(X) = sum\_{i} P(X = x_i) \cdot x_i $<br/>
+Each of the four outcomes has a probability of $ \frac{1}{4} $, as the coins are fair and independent.<br/>
 Therefore, we have:
-begin{itemize}
-	<li> $ P(X = 1) = \frac{1}{4} $ (from the $(H, H)$ outcome)
+<ul>
+   <li> $ P(X = 1) = \frac{1}{4} $ (from the $(H, H)$ outcome)
    <li> $ P(X = 0) = \frac{1}{4} $ (from the $(H, T)$ outcome)
    <li> $ P(X = -1) = \frac{1}{2} $ (from the $(T, H)$ and $(T, T)$ outcomes combined)
-   end{itemize}
+</ul>
 
-4. **Substitute the Values into the Expected Value Formula:**
-   [
-   E(X) = 1 \cdot \frac{1}{4} + 0 \cdot \frac{1}{4} + (-1) \cdot \frac{1}{2}
-   ]
-   [
-   E(X) = \frac{1}{4} + 0 - \frac{1}{2}
-   ]
-   [
-   E(X) = \frac{1}{4} - \frac{2}{4}
-   ]
-   [
-   E(X) = -\frac{1}{4}
-   ]
+4. **Substitute the Values into the Expected Value Formula:**<br/>
+   $E(X) = 1 \cdot \frac{1}{4} + 0 \cdot \frac{1}{4} + (-1) \cdot \frac{1}{2}$<br/>
+   $E(X) = \frac{1}{4} + 0 - \frac{1}{2}$<br/>
+   $E(X) = \frac{1}{4} - \frac{2}{4}$<br/>
+   $E(X) = -\frac{1}{4}$
 
-Therefore, the expected value of $ X $ is $ boxed{-\frac{1}{4}} $.
+Therefore, the expected value of $ X $ is $ \boxed{-\frac{1}{4}} $.