update fa24 midterm q07 explanation

docs/fa24-midterm/index.html

@@ -700,6 +700,26 @@ <h1 class="title"> </h1>
 </header>
 <p><strong>Answer</strong>:
 <code>candy_bowl, 3, replace=False</code></p>
+<p>The question asks us to “take 3 candies from the bowl at random.” In
+this part, we need to sample 3 candies at random using
+<code>np.random.choice</code>. Now, we evaluate each option one by one
+as follows:</p>
+<ul>
+<li><p><code>candy_bowl, len(candy_bowl), replace=False</code>: The code
+tries to sample all candies without replacement. However, we are asked
+to only sample three candies, not all.</p></li>
+<li><p><code>candy_bowl, 3, replace=False</code>: The code samples three
+candies without replacement, which matches the description. This option
+is correct.</p></li>
+<li><p><code>candy_bowl, 3, replace=True</code>: The code samples three
+candies from the bowl with replacement. Under this setting, the same
+candy can be selected multiple times in a single grab, which is not
+realistic.</p></li>
+<li><p><code>candy_bowl, repetitions, replace=True</code>: This option
+attempts to sample <code>repetitions</code> (10,000) candies in a single
+grab. We are asked to sample three candies per iteration of the loop,
+not thousands.</p></li>
+</ul>
 <hr/>
 <h5>Difficulty: ⭐️⭐️</h5>
 <p>
@@ -733,6 +753,31 @@ <h1 class="title"> </h1>
 </header>
 <p><strong>Answer</strong>:
 <code>grab[0] == grab[1] and grab[0] == grab[2]</code></p>
+<p>Here, we need condition that checks if all three candies selected in
+the grab are the same. We now analyze each option as follows:</p>
+<ul>
+<li><p><code>grab[0] == "Rolo" and grab[1] == "Rolo" and grab[2] == "Rolo"</code>:
+This condition explicitly checks if all three candies are “Rolo”. While
+it ensures that the three candies are the same, it only works for “Rolo”
+and not for other types of candy in the bowl (e.g., “Twix,”
+“M&amp;M”).</p></li>
+<li><p><code>grab[0] == grab[1] and grab[0] == grab[2]</code>: This
+condition checks if the first candy (grab[0]) is the same as the second
+(grab[1]) and the third (grab[2]). If all three candies are the same
+type (regardless of which type), this condition will evaluate to True.
+Otherwise, the expression will evaluate to False, which is what we need.
+The option is correct.</p></li>
+<li><p><code>grab[0] == grab[1] or grab[0] == grab[2]</code>: This
+condition checks if the first candy (grab[0]) matches either the second
+(grab[1]) or the third (grab[2]). It does not require all three candies
+to be the same. For example, if grab = [“Twix”, “Twix”, “M&amp;M”], this
+condition would incorrectly evaluate to True.</p></li>
+<li><p><code>grab == "Rolo" | grab == "M&amp;M"</code>: This condition
+is syntactically invalid. It tries to compare the grab list (which
+contains three elements) with two strings (“Rolo” and “M&amp;M”) using a
+bitwise OR (|), not to mention that it does not check if three candies
+are the same.</p></li>
+</ul>
 <hr/>
 <h5>Difficulty: ⭐️</h5>
 <p>
@@ -765,6 +810,19 @@ <h2 class="accordion-header" id="heading7_3">
 <h1 class="title"> </h1>
 </header>
 <p><strong>Answer</strong>: <code>prob_all_same / repetitions</code></p>
+<p>To calculate the estimated probability of drawing three candies of
+the same type, we divide the total number of successes
+(<code>prob_all_same</code>, which counts the instances where all three
+candies are identical) by the total number of iterations
+(<code>repetitions</code>).</p>
+<p>The option <code>prob_all_same.mean()</code> is incorrect because
+<code>prob_all_same</code> is an integer that accumulates the count of
+successful trials, not an array or list that supports the
+<code>.mean()</code> method. Similarly, dividing by
+<code>len(candy_bowl)</code> or <code>3</code> is incorrect, as neither
+represents the total number of iterations. Therefore, using these values
+as the denominator would not provide an accurate probability
+estimate.</p>
 <hr/>
 <h5>Difficulty: ⭐️⭐️</h5>
 <p>

problems/fa24-midterm/q07.md

@@ -31,6 +31,16 @@ What goes in blank (a)?
 
 **Answer**: `candy_bowl, 3, replace=False`
 
+The question asks us to "take 3 candies from the bowl at random." In this part, we need to sample 3 candies at random using `np.random.choice`. Now, we evaluate each option one by one as follows:
+
+- `candy_bowl, len(candy_bowl), replace=False`: The code tries to sample all candies without replacement. However, we are asked to only sample three candies, not all.
+
+- `candy_bowl, 3, replace=False`: The code samples three candies without replacement, which matches the description. This option is correct.
+
+- `candy_bowl, 3, replace=True`: The code samples three candies from the bowl with replacement. Under this setting, the same candy can be selected multiple times in a single grab, which is not realistic.
+
+- `candy_bowl, repetitions, replace=True`: This option attempts to sample `repetitions` (10,000) candies in a single grab. We are asked to sample three candies per iteration of the loop, not thousands.
+
 <average>88</average>
 
 # END SOLUTION
@@ -50,6 +60,17 @@ What goes in blank (b)?
 
 **Answer**: `grab[0] == grab[1] and grab[0] == grab[2]`
 
+Here, we need condition that checks if all three candies selected in the grab are the same. We now analyze each option as follows:
+
+- `grab[0] == "Rolo" and grab[1] == "Rolo" and grab[2] == "Rolo"`: This condition explicitly checks if all three candies are "Rolo". While it ensures that the three candies are the same, it only works for "Rolo" and not for other types of candy in the bowl (e.g., "Twix," "M&M").
+
+- `grab[0] == grab[1] and grab[0] == grab[2]`: This condition checks if the first candy (grab[0]) is the same as the second (grab[1]) and the third (grab[2]). If all three candies are the same type (regardless of which type), this condition will evaluate to True. Otherwise, the expression will evaluate to False, which is what we need. The option is correct.
+
+- `grab[0] == grab[1] or grab[0] == grab[2]`: This condition checks if the first candy (grab[0]) matches either the second (grab[1]) or the third (grab[2]). It does not require all three candies to be the same. For example, if grab = ["Twix", "Twix", "M&M"], this condition would incorrectly evaluate to True.
+
+- `grab == "Rolo" | grab == "M&M"`: This condition is syntactically invalid. It tries to compare the grab list (which contains three elements) with two strings ("Rolo" and "M&M") using a bitwise OR (|), not to mention that it does not check if three candies are the same.
+
+
 <average>92</average>
 
 # END SOLUTION
@@ -69,6 +90,10 @@ What goes in blank (c)?
 
 **Answer**: `prob_all_same / repetitions`
 
+To calculate the estimated probability of drawing three candies of the same type, we divide the total number of successes (`prob_all_same`, which counts the instances where all three candies are identical) by the total number of iterations (`repetitions`).
+
+The option `prob_all_same.mean()` is incorrect because `prob_all_same` is an integer that accumulates the count of successful trials, not an array or list that supports the `.mean()` method. Similarly, dividing by `len(candy_bowl)` or `3` is incorrect, as neither represents the total number of iterations. Therefore, using these values as the denominator would not provide an accurate probability estimate.
+
 <average>86</average>
 
 # END SOLUTION