Answer:

    orders.loc[orders["driver"] == "WOLVAA01", "minutes"].sort_values().iloc[1] 

or

    orders.loc[orders["driver"] == "WOLVAA01", "minutes"].sort_values(ascending=False).iloc[-2]

loc helps us extract a subset of the rows and columns in a DataFrame. In general, loc works as follows:

    orders.loc[<subset of rows>, <subset of columns>]

Here:

• We want all of the rows where orders["driver"] is equal to "WOLVAA01", so we use the expression orders["driver"] == "WOLVAA01" as the first input to loc. (Remember, orders["driver"] == "WOLVAA01" is a Series of Trues and Falses; only the True values are kept in the result.)

• We want just the column "minutes", so we supply it as a string as the second input to loc.

Now, we have a Series with the minutes of all orders that "WOLVAA01" has reviewed; calling the .sort_values() method on this Series will give us back another Series with the minutes of the orders sorted by default in ascending order. Since we are looking for the second fastest order, we can then use iloc to extract this value. As Python is zero-indexed this value will be 1.

Similarly, if you specify ascending=False as a parameter to .sort_values(), the orders will be sorted in descending order with the order taking the longest amount of time being on top. In such a case, we would want to extract the second last element from the series.

Answer:

• (i): "Driver". We use groupby("Driver") because we want to check the consistency of ratings for each individual driver. By grouping by "Driver", we are able to look at each driver’s set of ratings across their deliveries.

• (ii): "rating". We want to check whether all "rating" values for a given "Driver" are the same. To do so, we’ll need to look at a Series of "rating" values for each "Driver", so that’s the column we extract here.

• (iii): agg. The agg method allows us to apply an aggregation function to each group (each "Driver" in this case). This is necessary because we are checking a condition across an entire group of ratings rather than filtering individual rows. It’s also worth noting that since .mean() is used at the end, the agg step must be producing a series of Boolean values (True or False), one for each "Driver". Each value represents whether that driver’s ratings are consistent.

• (iv): x.max() == x.min() or x.nunique() == 1. This is the Boolean statement used in the f function to check whether a "Driver"’s ratings are consistent. x.max() == x.min() confirms that all ratings are identical, since both the maximum and minimum must be the same if no variation exists. x.nunique() == 1 achieves the same goal by checking directly whether there is only one unique rating. Either approach is valid, and there are several other possibilities too, such as checking x.std() == 0 or using x.eq(x.iloc[0]).all().

Answer: Every driver made exactly 0 or exactly k "Snack" deliveries, where k is some positive constant.

The expression runs a query on the orders DataFrame, selecting only rows where the "type" is equal to "Snack". This narrows the focus to the set of deliveries that were for "Snack" items. The code then selects the "driver" column to identify which driver handled each "Snack" delivery.

The first .value_counts() computes how many "Snack" deliveries each "driver" made. The second .value_counts() then counts how many drivers made each possible number of "Snack" deliveries.

Finally, .shape[0] measures how many distinct numbers of "Snack" deliveries were made by drivers. Since the expression evaluates to 1, we can conclude that there is only one distinct number of "Snack" deliveries made by drivers. This implies that every driver either made exactly 0 or exactly k "Snack" deliveries.

Answer: The driver who has driven the most minutes total, among all "Driver"s with at least 10 deliveries and a minimum review rating of 4.5.

The expression first groups the data by the "Driver" column. It then filters the groups to only include those where the "Driver" has made at least 10 deliveries and has a minimum "rating" of 4.5. Only "Driver"s who meet both conditions are kept in the output.

After filtering, the data is grouped by "Driver" again, and the total number of "minutes" each "Driver" has driven is calculated. Finally, .idxmax() identifies the "Driver" who has driven the most "minutes" among all "Driver"s that fit the two criteria above.

Answer: There are multiple possible answers.

• (i): 3 or -1. The index of C is sorted alphabetically, meaning the row order follows "Buffet", "Drinks", "Meal", "Snack". Since "Snack" is the last item in this order, its index is 3 (0-based) or -1 (negative indexing).

• (ii): 2 or -2. Similarly, the columns are sorted alphabetically: "High", "Low", "Moderate", "Very Low". "Moderate" appears third in this order, making its index 2 (0-based) or -2 (negative indexing).

• (iii): "Snack". To get the total number of "Snack" deliveries across all traffic levels, we use C.loc["Snack"]. The .sum() function then adds up all values in that row.

Answer: - (i): "Buffet". When using .loc, we reference labels instead of numerical indices. Since the index represents order "type" and is sorted alphabetically ("Buffet", "Drinks", "Meal", "Snack"), "Buffet" is the correct label for selecting the row corresponding to buffet orders.

• (ii): "Low". Similarly, the column order is sorted alphabetically as "High", "Low", "Moderate", "Very Low", making "Low" the correct column label to select deliveries in "Low" traffic.

• (iii): C.iloc[:, 1]. Here, we are using .iloc, which is integer-location based indexing. The [:, 1] part means we are selecting all rows (:) in the second column (index 1), which corresponds to "Low" traffic, since the columns are sorted alphabetically. By selecting C.iloc[:, 1], we are referencing all the values in the "Low" column, and .sum() will give the total number of deliveries made in "Low" traffic across all order types.

Answer: "Drinks"

Given that the resulting DataFrame after the merge has 7 rows, this means that there are 7 matches between the rows in A and B based on the "type" column. The value in the third row of DataFrame B is unknown. We need to figure out which value for "type" will give exactly 7 matches when merged with A.

Since merging on "type" generates rows where both DataFrames have the same "type", we need to look at the counts of each type in A and B to deduce which value the unknown "type" should be. We can see that there are 5 rows with type "Buffet" in DataFrame A and 1 row with type "Buffet" in DataFrame B. This means in our merged dataset we will have at least 5 rows of "Buffet".

We can then go through each of the different options to see which option will give us exactly 7 rows in our merged dataset. If the unknown value is:

• Drinks: We have 2 rows with type "Drinks" in DataFrame A and 1 row with type "Drinks" (unknown value replaced) in DataFrame B, which gives us 2 \times 1 = 2 rows in our merged dataset. This is the correct answer because now we will have 5 (Buffet rows) + 2 (Drinks rows) = 7 rows in the merged dataset.

• Buffet: We have 5 rows of "Buffet" in DataFrame A and 2 rows in DataFrame B, which gives us 2 \times 5 = 10 rows in our merged dataset.

• Meal: We have 0 rows of "Meal" in DataFrame A and 2 row in DataFrame B, which gives us 0 \times 2 = 0 row in our merged dataset.

• Snack: We have 0 rows of "Snack" in DataFrame A and 3 rows in DataFrame B, which gives us 0 \times 3 = 0 rows in our merged dataset. This is the correct answer because now we will have exactly 10 rows (all Buffet rows) in total.

Answer: "Buffet"

Using the same logic as in the solution to 6.1, if the unknown value is "Buffet", the merged dataset will include:

• 5 (Buffet rows from A) * 2 (Buffet rows from B) = 10 rows.

This matches the stated total, so the unknown must be "Buffet".

Answer: 14

We can replace the unknown value with each type and compute how many rows we expect to see in the final dataset after an outer merge:

• If the unknown value is Drinks: We will have 2 (Drinks rows) + 5 (Buffet rows) + 2 (Meal rows) + 2 (Snack rows) = 11 rows.
• If the unknown value is Buffet: We will have 10 (Buffet rows) + 2 (Drink rows from DataFrame A) + 2 (Meal rows) + 2 (Snack rows) = 16 rows.
• If the unknown value is Meal: We will have 3 (Meal rows) + 5 (Buffet rows) + 2 (Drink rows) + 2 (Snack rows) = 12 rows.
• If the unknown value is Snack: We will have 3 (Snack rows) + 5 (Buffet rows) + 2 (Meal rows) + 2 (Meal rows) = 12 rows.

Thus, the only possible number of rows that cannot occur is 14.

Answer: 2

When merging on both "type" and "rating", only exact matches in both columns will contribute to the result. Since the unknown value is "Buffet", the only matching combination is "Buffet" with rating 4.6, which appears once in each DataFrame.

Thus, the total number of rows is 1 (from A) * 1 (from B) = 1 per matching pair, and if both datasets hold one each, the result has 2 rows.

Correct answers:

• one_sim draws k drivers with replacement, instead of without replacement
• simulation only picks one value of k, when it should select a new k on each iteration

one_sim draws k drivers with replacement, instead of without replacement

Because np.random.choice() is called separately each time within the for loop, there is no guarantee that all k chosen drivers will be distinct across iterations. Which means that some drivers might be picked more than once, violating the “unique drivers” requirement. The correct way to choose the k drivers would be -

np.random.choice(drivers, k, replace=False)

simulation only picks one value of k, when it should select a new k on each iteration

The code defines k once at the beginning of simulation() using choose_k(). This means that the same value of k is used for all iterations of the simulation. However, as per the problem statement, k should be chosen anew on each iteration of the simulation. Therefore, the function choose_k() should be called within the loop to select a random k for each iteration.

Now, for the other options:

• drivers only includes the drivers that made one delivery, rather than all drivers

  orders["driver"].unique() returns all unique drivers, regardless of how many deliveries they made, so this issue doesn’t exist in the code.

• choose_k returns a random integer between 1 and 11, instead of one from 1 to 10

  np.random.choice(np.arange(1, 11)) correctly selects a random integer from 1 to 10 (inclusive), so this is not an issue.

• choose_k always returns the same number

  choose_k() generates a random number between 1 and 10 every time it is called. Therefore, it does not always return the same number.

• one_sim always returns False, because selected is always an empty array

  The array selected is updated correctly inside the loop, and it will contain the selected drivers for the simulation, so it doesn’t always return False.

• one_sim draws k drivers without replacement, instead of with replacement

  The issue is not about drawing without replacement; the issue is that the function draws with replacement within the for loop, which could result in the same driver being chosen more than once.

D.shape[0] is guaranteed to evaluate to the same value before and after the imputation code is run because it represents the total number of rows in the DataFrame, which is unaffected by filling in missing values. The imputation modifies the “minutes” column but does not alter the number of rows.

Answer: D["minutes"].mean(), D.loc[D["traffic"] == "Moderate", "minutes"].mean()

In this imputation code we are performing mean imputation by filling the missing values with the mean of the whole dataset.

D["minutes"].mean() - As we’ve seen in class, replacing missing values with the mean of the observed values does not change the mean.

D.loc[D["traffic"] == "Moderate", "minutes"].mean() - Since the imputation does not affect the rows where “traffic” is “Moderate” (i.e., these rows already have no missing values), the mean for this subset of the data will remain unchanged as well.

D.groupby("traffic")["minutes"].mean() - This expression will change after imputation. The grouped means for traffic categories that had missing values will be affected by the imputation, as we’re replacing missing values with the overall mean rather than group-specific means.

Answer: D.loc[D["traffic"] == "Moderate", "minutes"].mean(), D.groupby("traffic")["minutes"].mean()

In this imputation code we are performing conditional mean imputation by filling the missing values with the minutes of each group of traffic.

D["minutes"].mean() - This expression will change after imputation. When we fill missing values with group-specific means, the overall mean can change.

D.loc[D["traffic"] == "Moderate", "minutes"].mean() - Since the imputation only fills in missing values for the “minutes” column based on the mean of the “minutes” values within each traffic group, the mean for “Moderate” traffic will remain unchanged.

D.groupby("traffic")["minutes"].mean() - The imputation groups by the “traffic” column and fills missing values using the mean of “minutes” within each group. Since “Moderate” traffic already had no missing values, the mean for this group will not be affected, and neither will the means for other groups, assuming the missing values are only filled within their respective groups.

Answer: D.loc[D["traffic"] == "Moderate", "minutes"].mean()

In this imputation code we are performing unconditional probabilistic imputation by filling the missing values by randomly sampling from the present values in the “minutes” column.

D["minutes"].mean() - This expression will likely change after imputation, as we’re replacing missing values with randomly selected observed values.

D.loc[D["traffic"] == "Moderate", "minutes"].mean() - Since “Moderate” traffic rows have no missing values, the imputation does not affect them. Therefore, the mean for this subset of data will remain unchanged.

D.groupby("traffic")["minutes"].mean() - The means for traffic categories that had missing values will likely change after imputation, as we’re filling in those missing values with randomly selected values from the entire dataset rather than values specific to each traffic category.

Answer:

Answer: 5

The bag of words vector for Document 1 is: \begin{bmatrix} 1 \\ 1 \\ k \\ 0 \\ 0 \end{bmatrix}

The bag of words vector for Document 2 is:\begin{bmatrix} 0 \\ 0 \\ 1 \\ 1 \\ 1 \end{bmatrix}

The cosine similarity can be calculated using the formula: \text{cosine sim}(\vec d_1, \vec d_2) = \frac{\vec d_1 \cdot \vec d_2}{\lVert d_1 \rVert \lVert d_2 \rVert}

For the two vectors \vec{d_1} = (1, 1, k, 0, 0) and \vec{d_2} = (0, 0, 1, 1, 1), the dot product is: \text{Dot product} = (1)(0) + (1)(0) + (k)(1) + (0)(1) + (0)(1) = k

Next, we need to calculate the magnitudes (or norms) of the two vectors. \|\vec{d_1}\| = \sqrt{1^2 + 1^2 + k^2 + 0^2 + 0^2} = \sqrt{2 + k^2}

\|\vec{d_2}\| = \sqrt{0^2 + 0^2 + 1^2 + 1^2 + 1^2} = \sqrt{3}

Substitute the values we calculated: \text{Cosine similarity} = \frac{k}{\sqrt{2 + k^2} \cdot \sqrt{3}}

We are given that the cosine similarity is \frac{5}{9}. So, we set up the equation: \frac{k}{\sqrt{2 + k^2} \cdot \sqrt{3}} = \frac{5}{9}

Solving for k gives us 5.

Answer:

• butter and chicken

• curry and with

The TF-IDF score of a term in a document is given by: \text{tf-idf}(t,d) = \text{tf}(t,d) \cdot \text{idf}(t) = \frac{\# \text{ of occurrences of } t \text{ in } d }{\text{total } \# \text{ of terms in } d} \cdot \log_2\left(\frac{\text{total } \# \text{ of documents}}{\# \text{ of documents in which } t \text{ appears}}\right)

TF-IDF scores are high when:

1. A term occurs frequently in the document (high TF), and

2. A term appears in few documents in the corpus (high IDF).

Naan appears in both documents, so its IDF is 0 and its TF-IDF is 0 in both.

Butter and chicken each appear only once, and only in document 1. So their TF is low, but their IDF is high.

Curry and with each appear only once, and only in document 2. So they also have high IDF and non-zero TF.

Therefore, butter and chicken have the highest TF-IDF in document 1, and curry and with in document 2.

Answer: \text{WS}(t, d) < 1

The Wolverine Score compares a term’s relative frequency in document d versus its average frequency in the entire corpus.

The first part of WS, \frac{\text{total terms in } d}{\# \text{ of occurrences of } t \text{ in } d}, is the inverse of how frequent t is in d.

The second part, \frac{\text{total occurrences of } t \text{ in corpus}}{\text{total terms in corpus}}, is the overall frequency of t.

If WS is less than 1, it means that the term occurs more often within d than on average across all documents. This implies that the term is especially important or relevant to d.

Example: If a term appears 5 times in a 10-word document (50%), but only 10 times in a 100-word corpus (10%), then: \text{WS} = \left(\frac{10}{5}\right) \cdot \left(\frac{10}{100}\right) = 2 \cdot 0.1 = 0.2 < 1 So WS < 1 means the term is concentrated in d.

Answer:

• soup.find_all("div", class_="menu-item") or soup.find_all("div", attrs={"class": "menu-item"}). 

• float(x.get("data-calories")) < 500

• x.find("h2").text 

soup.find_all("div", class_="menu-item") or soup.find_all("div", attrs={"class": "menu-item"}). 

We want to extract all the menu items from the HTML. Each menu item is contained within a div tag with the class menu-item. So, we need to find all the div elements that have this class. In BeautifulSoup, we can do this with the find_all() method, which searches the document for all elements matching a specific condition.

float(x.get("data-calories")) < 500 

This code is used to check whether the number of calories in each menu item is less than 500. We first use x.get("data-calories") to retrieve the calorie count as a string. We then convert that string to a floating-point number using float() so we can compare it with 500. If the calorie count is less than 500, we extract the name of the menu item from the <h2> tag and add it to the low_cals list.

x.find("h2").text

Lastly, we want to extract the name of the menu item from each selected div element. The find(“h2”) method is called on x. .find searches for the first occurrence of a <h2> tag within the x element. Once we have the <h2> element (in this case, Chana Masala), we use .text to extract the text content inside the <h2> tag.

Answer:

• (\d+)\.(\d{3})

• int(p[0] + p[1]) / 100 

(\d+)\.(\d{3})

(\d+): This part matches one or more digits before the first decimal point. The parentheses around capture this part into a group, allowing us to extract the digits before the first decimal.

\.: This matches the literal decimal point. The backslash  is necessary because the dot . is a special character in regular expressions that usually matches any character, so we need to escape it with a backslash to match an actual period.

(\d{3}): This matches exactly three digits after the first decimal and before the second decimal. The parentheses around capture these digits into another group, allowing us to extract them separately.

int(p[0] + p[1]) / 100 

p is a tuple returned by the split_pieces(s) function, where p[0] is the first part (before the first decimal point) and p[1] is the part between the two decimals.

For example, if the input string is “8.334.108”, p would be (“8”, “334”). The expression python p[0] + p[1] concatenates these two parts together, creating the string “83334”. This represents the combined value of the minutes, but it’s not in the correct format. We can wrap this around the int function which will convert the concatenated string “83334” into an integer.

Answer: \bar{y}

To find the optimal parameter, we have to minimize the average loss function. The loss function is given as: L_{\alpha \beta}(y_i, h) = (3y_i - 3h)^2

To minimize this, we need to find the value of h that minimizes the sum of squared errors for all y_{i}. We do this by taking the derivative of the loss function and solving it by setting it to zero.

Expanding the squared term: \sum_{i=1}^{n} (9y_i^2 - 18y_i h + 9h^2)

Using summation properties: 9 \sum_{i=1}^{n} y_i^2 - 18h \sum_{i=1}^{n} y_i + 9h^2 n

Differentiating: \frac{d}{dh} \left( 9 \sum_{i=1}^{n} y_i^2 - 18h \sum_{i=1}^{n} y_i + 9h^2 n \right)

Since the first term 9 \sum_{i=1}^{n} y_i^2 does not depend on h, its derivative is zero.

Differentiating the remaining terms: \frac{d}{dh} (-18h \sum_{i=1}^{n} y_i) = -18 \sum_{i=1}^{n} y_i

\frac{d}{dh} (9h^2 n) = 18h n

Thus, the derivative of the total loss is: -18 \sum_{i=1}^{n} y_i + 18h n

Setting the derivative equal to zero: -18 \sum_{i=1}^{n} y_i + 18h n = 0

Dividing both sides by 18: - \sum_{i=1}^{n} y_i + h n = 0

Solving for h: h = \frac{\sum_{i=1}^{n} y_i}{n}

This is equivalent to mean of \bar{y}. Thus, the optimal constant prediction is the mean of the delivery times.

Answer: 2 \bar{y}

To find the optimal parameter, we have to minimize the average loss function. The loss function is given as:
L_{\alpha \beta}(y_i, h) = (6y_i - 3h)^2

To minimize this, we need to find the value of h that minimizes the sum of squared errors for all y_i. We do this by taking the derivative of the loss function and solving it by setting it to zero.

Expanding the squared term: \sum_{i=1}^{n} (36y_i^2 - 36y_i h + 9h^2)

Using summation properties: 36 \sum_{i=1}^{n} y_i^2 - 36h \sum_{i=1}^{n} y_i + 9h^2 n

Differentiating: \frac{d}{dh} \left( 36 \sum_{i=1}^{n} y_i^2 - 36h \sum_{i=1}^{n} y_i + 9h^2 n \right)

Since the first term 36 \sum_{i=1}^{n} y_i^2 does not depend on h, its derivative is zero.

Differentiating the remaining terms: \frac{d}{dh} (-36h \sum_{i=1}^{n} y_i) = -36 \sum_{i=1}^{n} y_i

\frac{d}{dh} (9h^2 n) = 18h n

Thus, the derivative of the total loss is: -36 \sum_{i=1}^{n} y_i + 18h n

Setting the derivative equal to zero: -36 \sum_{i=1}^{n} y_i + 18h n = 0

Dividing both sides by 18: -2 \sum_{i=1}^{n} y_i + h n = 0

Solving for h: h = \frac{2 \sum_{i=1}^{n} y_i}{n} = 2 \bar{y}

Thus, the optimal constant prediction is 2 \bar{y}.

Answer: \frac{\alpha}{\beta} \bar{y}

To find the optimal parameter, we have to minimize the average loss function. The loss function is given as:
L_{\alpha \beta}(y_i, h) = (\alpha y_i - \beta h)^2

To minimize this, we need to find the value of h that minimizes the sum of squared errors for all y_i. We do this by taking the derivative of the loss function and solving it by setting it to zero.

Expanding the squared term: \sum_{i=1}^{n} (\alpha^2 y_i^2 - 2\alpha\beta y_i h + \beta^2 h^2)

Using summation properties: \alpha^2 \sum_{i=1}^{n} y_i^2 - 2\alpha\beta h \sum_{i=1}^{n} y_i + \beta^2 h^2 n

Differentiating: \frac{d}{dh} \left( \alpha^2 \sum_{i=1}^{n} y_i^2 - 2\alpha\beta h \sum_{i=1}^{n} y_i + \beta^2 h^2 n \right)

Since the first term \alpha^2 \sum_{i=1}^{n} y_i^2 does not depend on h, its derivative is zero.

Differentiating the remaining terms: \frac{d}{dh} (-2\alpha\beta h \sum_{i=1}^{n} y_i) = -2\alpha\beta \sum_{i=1}^{n} y_i

\frac{d}{dh} (\beta^2 h^2 n) = 2\beta^2 h n

Thus, the derivative of the total loss is: -2\alpha\beta \sum_{i=1}^{n} y_i + 2\beta^2 h n

Setting the derivative equal to zero: -2\alpha\beta \sum_{i=1}^{n} y_i + 2\beta^2 h n = 0

Dividing both sides by 2: - \alpha\beta \sum_{i=1}^{n} y_i + \beta^2 h n = 0

Solving for h: h = \frac{\alpha\beta \sum_{i=1}^{n} y_i}{\beta^2 n}

We substitute \sum_{i=1}^{n} y_i = n\bar{y} into the equation: h = \frac{\alpha\beta (n\bar{y})}{\beta^2 n}

This simplified gives us h^* = \frac{\alpha}{\beta} \bar{y}

Answer: >

The quantity \frac{1}{n} \sum_{i = 1}^n (y_i - F(x_i))^2 is the mean squared error (MSE) of model F. By definition, F is the line that minimizes MSE over all possible linear models.

The quantity \frac{1}{n} \sum_{i = 1}^n (y_i - G(x_i))^2 is the mean squared error of model G, which is not optimized for MSE but rather for mean absolute error (MAE).

Since F is specifically constructed to minimize the squared error, and G is not, we must have:
\sum_{i = 1}^{n} (y_i - G(x_i))^2 > \sum_{i = 1}^{n} (y_i - F(x_i))^2

Answer: <

The quantity \frac{1}{n} \sum_{i = 1}^n | y_i - G(x_i) | is the mean absolute error (MAE) of model G. By definition, G is the line that minimizes MAE over all possible linear models.

The quantity \frac{1}{n} \sum_{i = 1}^n | y_i - F(x_i) | is the mean absolute error of model F, which is not optimized for MAE but rather for mean squared error (MSE).

Since G is specifically constructed to minimize the absolute error, and F is not, we must have:
\sum_{i = 1}^{n} | y_i - G(x_i) | < \sum_{i = 1}^{n} | y_i - F(x_i) |

Squaring both sides preserves the inequality (since both sides are non-negative), so:
\left( \sum_{i = 1}^{n} | y_i - G(x_i) | \right)^2 < \left( \sum_{i = 1}^{n} | y_i - F(x_i) | \right)^2

Answer: Line 1

The key idea is that models trained with squared loss (MSE) are more sensitive to outliers than models trained with absolute loss (MAE).

Since Line 1 appears to be “pulled up” more strongly by an outlier, it suggests that this line was influenced more heavily by extreme values. This behavior aligns with how MSE-based regression works: outliers have a greater impact on the overall loss because squaring the errors makes large deviations even more significant.

In contrast, MAE-based regression (Line 2) is less sensitive to outliers because absolute differences do not grow as quickly.

Therefore, Line 1 corresponds to F, the MSE-minimizing line.

Answer: \frac{55}{100}

We have 100 unique drivers and 10 different groups. The probability that "WOLVAA01" wins is calculated by adding up the chances that it is chosen in each group.

For each group, the probability of winning depends on how many drivers are in that group. For example:

• In the first group, the probability that "WOLVAA01" wins is \frac{1}{10} \times \frac{1}{100}.
• In the second group, the probability is \frac{1}{10} \times \frac{2}{100}.
• This continues until the tenth group, where the probability is \frac{1}{10} \times \frac{10}{100}.

The total probability is the sum of these individual probabilities: \text{Total Probability} = \frac{1}{10} \times \frac{1}{100} + \frac{1}{10} \times \frac{2}{100} + \cdots + \frac{1}{10} \times \frac{10}{100} which simplifies to: \text{Total Probability} = \frac{1}{1000} (1 + 2 + \cdots + 10) = \frac{1}{1000} \times 55 = \frac{55}{1000} = \frac{55}{100}.