The following table summarizes the ice cream flavor and topping choices
  made by a group of people at an ice cream social.

Number of people who had each:	B = Bananas	C = Chocolate Chips	TOTAL
S = Strawberry	   22	1e)_________	   46
V = Vanilla	1c)_________	1d)_________	1a)_________
  TOTAL	1b)_________	   39	   72

Let T = {The set of all people who ate a bowl of ice cream at the social}

#1. Complete the table by answering the questions IN THE ORDER THAT THEY
  ARE SHOWN IN THE TABLE.

a) How many people chose vanilla ice cream? 
26
Correct
b) How many people had bananas on top of their ice cream? 
33
Correct
c) How many people chose vanilla ice cream with bananas on top? 
11
Correct
d) How many people chose vanilla ice cream with chocolate chips on top? 
15
Correct
e) How many people chose strawberry ice cream with chocolate chips on top? 
24
Correct



#2. FOR EACH QUESTION BELOW:

* Select the correct formula to use, for the corresponding question,
  from the list shown below.

* Find and round results to the nearest one-hundredth of a percent.

A.  
P
(
S
)
=
n
(
S
)
n
(
T
)
    B.  
P
(
V
)
=
n
(
V
)
n
(
T
)
    C.  
P
(
B
)
=
n
(
B
)
n
(
T
)
    D.  
P
(
C
)
=
n
(
C
)
n
(
T
)
 


E.  
P
(
C
∣
S
)
=
n
(
C
∩
S
)
n
(
S
)
    F.  
P
(
S
∣
C
)
=
n
(
S
∩
C
)
n
(
C
)
   

G.  
P
(
B
∣
S
)
=
n
(
B
∩
S
)
n
(
S
)
   H.  
P
(
S
∣
B
)
=
n
(
S
∩
B
)
n
(
B
)
    I.  
P
(
T
)
=
1
 


J.  
P
(
C
∣
V
)
=
n
(
C
∩
V
)
n
(
V
)
   K.  
P
(
V
∣
C
)
=
n
(
V
∩
C
)
n
(
C
)
 


L.  
P
(
V
∣
B
)
=
n
(
V
∩
B
)
n
(
B
)
    M.  
P
(
B
∣
V
)
=
n
(
B
∩
V
)
n
(
V
)
 




columns

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

### Completing the table:

We are given the following partial table:

| **Flavor/Topping Choices**     | **B = Bananas** | **C = Chocolate Chips** | **Total** |
|---------------------------------|----------------|-------------------------|-----------|
| **S = Strawberry**              | 22             | \( 1e) \)              | 46        |
| **V = Vanilla**                 | \( 1c) \)      | \( 1d) \)              | \( 1a) \) |
| **Total**                       | \( 1b) \)      | 39                      | 72        |

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:

| **Flavor/Topping Choices**     | **B = Bananas** | **C = Chocolate Chips** | **Total** |
|---------------------------------|----------------|-------------------------|-----------|
| **S = Strawberry**              | 22             | 24                      | 46        |
| **V = Vanilla**                 | 11             | 15                      | 26        |
| **Total**                       | 33             | 39                      | 72        |

### 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.

The following table summarizes the ice cream flavor and topping choices made by a group of people at an ice cream social.
Number of people who had each: B = Bananas, C = Chocolate Chips, TOTAL
S = Strawberry, 22, ___, 46
V = Vanilla, ___, ___, ___
TOTAL, ___, 39, 72
Let T = {The set of all people who ate a bowl of ice cream at the social}
Complete the table and answer questions on probability.

Solve a probability problem based on an ice cream social, where participants chose different flavors and toppings. Learn how to complete the table and calculate probabilities for vanilla, strawberry, bananas, and chocolate chips, using key formulas and concepts. Suitable for high school students in Grades 9-12.

Math Problem Statement

A.
P ( S )

n ( S ) n ( T ) B.
P ( V )

n ( V ) n ( T ) C.
P ( B )

n ( B ) n ( T ) D.
P ( C )

E.
P ( C ∣ S )

n ( C ∩ S ) n ( S ) F.
P ( S ∣ C )

G.
P ( B ∣ S )

n ( B ∩ S ) n ( S ) H.
P ( S ∣ B )

n ( S ∩ B ) n ( B ) I.
P ( T )

J.
P ( C ∣ V )

n ( C ∩ V ) n ( V ) K.
P ( V ∣ C )

L.
P ( V ∣ B )

n ( V ∩ B ) n ( B ) M.
P ( B ∣ V )

Solution

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Math Problem Statement

A. P ( S )

n ( S ) n ( T ) B. P ( V )

n ( V ) n ( T ) C. P ( B )

n ( B ) n ( T ) D. P ( C )

E. P ( C ∣ S )

n ( C ∩ S ) n ( S ) F. P ( S ∣ C )

G. P ( B ∣ S )

n ( B ∩ S ) n ( S ) H. P ( S ∣ B )

n ( S ∩ B ) n ( B ) I. P ( T )

J. P ( C ∣ V )

n ( C ∩ V ) n ( V ) K. P ( V ∣ C )

L. P ( V ∣ B )

n ( V ∩ B ) n ( B ) M. P ( B ∣ V )