From 2d3e7d69cc0787b7bbbdff57ac63c7540334806c Mon Sep 17 00:00:00 2001
From: andyeisenberg <andyjeisenberg@gmail.com>
Date: Thu, 29 Jan 2026 06:35:50 -0500
Subject: [PATCH] Adding some answers

---
 source/sec-Continuous-RVs.ptx      | 64 +++++++++++++++++++++++--
 source/sec-Discrete-RVs.ptx        | 19 ++++++++
 source/sec-Joint-Distributions.ptx | 77 +++++++++++++++++++++++++++++-
 3 files changed, 155 insertions(+), 5 deletions(-)

diff --git a/source/sec-Continuous-RVs.ptx b/source/sec-Continuous-RVs.ptx
index 161c649..b19ea2b 100644
--- a/source/sec-Continuous-RVs.ptx
+++ b/source/sec-Continuous-RVs.ptx
@@ -298,13 +298,42 @@
 
   <exercises xml:id="exercises-Continuous-RVs">
   <exercise>
-    <statement>
+    <introduction>
       <p>
         A continuous random variable <m>X</m> taking values in <m>[1, 4]</m> has p.d.f.
         <m>f(x) = k(x - \sqrt{x})</m> for some constant <m>k</m>.
-        What is the value of <m>k</m>?
       </p>
-    </statement>
+    </introduction>
+
+
+    <task>
+      <statement>
+        <p>
+          What is the value of <m>k</m>?
+        </p>
+      </statement>
+
+      <answer>
+        <p>
+          <m>k = \frac{6}{17}.</m>
+        </p>
+      </answer>
+    </task>
+
+
+    <task>
+      <statement>
+        <p>
+          Find <m>\Pr(2 \leq X \leq 3).</m>
+        </p>
+      </statement>
+
+      <answer>
+        <p>
+          <m>\Pr(2 \leq X \leq 3) \approx 0.325</m>
+        </p>
+      </answer>
+    </task>
   </exercise>
 
   <exercise>
@@ -318,6 +347,12 @@
         to find <m>\Pr\left(1 \leq X \leq \frac{3}{2}\right)</m>.
       </p>
     </statement>
+
+    <answer>
+      <p>
+        <m>\frac{1}{2}\left(\frac{x^2}{2} - \frac{1}{x}\right) + \frac{1}{4}</m>. <m>\Pr\left(1 \leq X \leq \frac{3}{2}\right) \approx 0.479.</m>
+      </p>
+    </answer>
   </exercise>
 
   <exercise>
@@ -329,8 +364,29 @@
         <m>f(x)</m>.
       </p>
     </statement>
+
+    <answer>
+      <p>
+        <m>f(x) = x^2 - 2x.</m>
+      </p>
+    </answer>
   </exercise>
 
+  <exercise>
+    <statement>
+      <p>
+        Let <m>T \sim \Exp(3)</m>.
+        Find <m>\Pr(1 \leq T \leq 3)</m> and <m>\Pr(T \geq 0.5)</m>.
+      </p>
+    </statement>
+
+    <answer>
+      <p>
+        <m>\Pr(1 \leq T \leq 3) \approx 0.050</m>. <m>\Pr(T \geq 0.5) \approx 0.223</m>.
+      </p>
+    </answer>
+  </exercise>
+  <!--
   <exercise>
     <statement>
       <p>
@@ -350,5 +406,5 @@
       </p>
     </statement>
   </exercise>
-  </exercises>
+  --> </exercises>
 </section>
\ No newline at end of file
diff --git a/source/sec-Discrete-RVs.ptx b/source/sec-Discrete-RVs.ptx
index 64fbaba..51c7689 100644
--- a/source/sec-Discrete-RVs.ptx
+++ b/source/sec-Discrete-RVs.ptx
@@ -398,6 +398,13 @@
           Which flip has the highest chance of being the first flip to come up heads?
         </p>
       </statement>
+
+      <solution>
+        <p>
+          <m>\Pr(T = k) = 0.6^{k-1} \cdot 0.4</m>, so every additional flip multiplies the probability by 0.6.
+          Therefore, the highest value for <m>\Pr(T = k)</m> occurs when <m>k = 1</m>, in which case <m>\Pr(T = 1) = 0.4.</m>
+        </p>
+      </solution>
     </task>
   </exercise>
   <!--
@@ -418,6 +425,18 @@
         During a 4-hour afternoon shift, what is the probability of serving 80 customers.
       </p>
     </statement>
+
+    <solution>
+      <p>
+        Let <m>N</m> be the number of customers seen during the afternoon shift.
+        Since the store averages 20 customers per hour, it will average 80 per 4-hours.
+        So <m>N \sim \Poiss(20, 4)</m> will have distribution <m>\Pr(N = k) = \frac{80^{k}}{k!} e^{-80}</m>.
+        Therefore:
+        <md>
+          <mrow> \Pr(N = 80) = \frac{80^{80}}{80!} e^{-80} \approx 0.045. </mrow>
+        </md>
+      </p>
+    </solution>
   </exercise>
   <!--
   <exercise>
diff --git a/source/sec-Joint-Distributions.ptx b/source/sec-Joint-Distributions.ptx
index 59276c6..91225e9 100644
--- a/source/sec-Joint-Distributions.ptx
+++ b/source/sec-Joint-Distributions.ptx
@@ -120,9 +120,84 @@
   <exercise>
     <statement>
       <p>
-        TODO
+        Consider <m>X, Y</m> with the joint distribution table below.
+        Are <m>X, Y</m> independent?
+      </p>
+
+      <table>
+        <title>Joint distribution for <m>X, Y</m></title>
+
+        <tabular halign="center">
+          <row bottom="minor">
+            <cell right="minor"></cell>
+            <cell right="minor"><m>X = 0</m></cell>
+            <cell><m>X = 1</m></cell>
+          </row>
+
+          <row bottom="minor">
+            <cell right="minor"><m>Y = 0</m></cell>
+            <cell right="minor">0.2</cell>
+            <cell>0.3</cell>
+          </row>
+
+          <row>
+            <cell right="minor"><m>Y = 1</m></cell>
+            <cell right="minor">0.4</cell>
+            <cell>0.1</cell>
+          </row>
+        </tabular>
+      </table>
+    </statement>
+
+    <answer>
+      <p>
+        No.
+        For example, <m>\Pr(X = 0, Y = 0) \neq \Pr(X = 0)\Pr(Y = 0).</m>
+      </p>
+    </answer>
+  </exercise>
+
+  <exercise>
+    <statement>
+      <p>
+        Suppose <m>X, Y</m> have the distributions:
+        <md>
+          <mrow> \Pr(X = 0) \amp = 0.1 \amp \Pr(Y = 0) \amp = 0.4 </mrow>
+          <mrow> \Pr(X = 1) \amp = 0.4 \amp \Pr(Y = 1) \amp = 0.6 </mrow>
+          <mrow> \Pr(X = 2) \amp = 0.5  </mrow>
+        </md>
+        Assuming <m>X, Y</m> are independent, write a joint distribution table.
       </p>
     </statement>
+
+    <answer>
+      <table>
+        <title>Joint Distribution</title>
+
+        <tabular halign="center">
+          <row bottom="minor">
+            <cell right="minor"></cell>
+            <cell right="minor"><m>X = 0</m></cell>
+            <cell right="minor"><m>X = 1</m></cell>
+            <cell><m>X = 2</m></cell>
+          </row>
+
+          <row bottom="minor">
+            <cell right="minor"><m>Y = 0</m></cell>
+            <cell right="minor">0.04</cell>
+            <cell right="minor">0.16</cell>
+            <cell>0.2</cell>
+          </row>
+
+          <row>
+            <cell right="minor"><m>Y = 1</m></cell>
+            <cell right="minor">0.06</cell>
+            <cell right="minor">0.24</cell>
+            <cell>0.3</cell>
+          </row>
+        </tabular>
+      </table>
+    </answer>
   </exercise>
   </exercises>
 </section>
\ No newline at end of file