Let's break down the problem and solve it step-by-step.

Completing the table:

We are given the following partial table:

We are given the values:

• $1a)$, total number of people who chose vanilla ice cream = 26
• $1b)$, total number of people who had bananas = 33
• $1c)$, number of people who chose vanilla ice cream with bananas = 11
• $1d)$, number of people who chose vanilla ice cream with chocolate chips = 15
• $1e)$, number of people who chose strawberry ice cream with chocolate chips = 24

We can now complete the table using this information.

1. Total for "V = Vanilla" column (i.e., $1a)$): 26
   We already know this value is provided.

2. Bananas total (row total) (i.e., $1b)$): 33
   This is given as the total number of people who had bananas.

3. Vanilla ice cream with bananas (i.e., $1c)$): 11
   This is given.

4. Vanilla ice cream with chocolate chips (i.e., $1d)$): 15
   This is given.

5. Strawberry ice cream with chocolate chips (i.e., $1e)$): 24
   This is given.

Now, we calculate the missing values:

Missing values:

• The total number of people who chose strawberry ice cream with bananas can be found by subtracting the chocolate chips total for strawberry from the total number of people who had strawberry ice cream: $22 = 46 - 24$

• The total number of people who chose bananas on vanilla ice cream is already given as 11. Therefore, the rest of the people who chose bananas had strawberry ice cream (i.e., 22).

Thus, the complete table is:

Probabilities:

Next, let's calculate the probabilities based on the formulas.

1. $P(S) = \frac{n(S)}{n(T)} = \frac{46}{72} \approx 0.64$ (A)
2. $P(V) = \frac{n(V)}{n(T)} = \frac{26}{72} \approx 0.36$ (B)
3. $P(B) = \frac{n(B)}{n(T)} = \frac{33}{72} \approx 0.46$ (C)
4. $P(C) = \frac{n(C)}{n(T)} = \frac{39}{72} \approx 0.54$ (D)

For the conditional probabilities:

5. $P(C|S) = \frac{n(C \cap S)}{n(S)} = \frac{24}{46} \approx 0.52$ (E)

Would you like to explore further details on any step? Here are some related questions to consider:

1. What is the probability that a randomly chosen person had both bananas and vanilla ice cream?
2. What is the probability that a randomly chosen person had both chocolate chips and vanilla ice cream?
3. What is the probability that a person who had bananas also had strawberry ice cream?
4. What is the probability that a person who had vanilla ice cream also had chocolate chips?
5. What is the probability that a person who had chocolate chips also had vanilla ice cream?

Tip: Conditional probabilities help in finding the likelihood of one event occurring given that another event has already happened.