To solve this problem, we need to analyze the given information and use set theory to find the answers to each question. Let's define:

• $n(A)$ = number of persons who used product A
• $n(B)$ = number of persons who used product B
• $n(C)$ = number of persons who used product C
• $n(A \cap B)$ = number of persons who used both products A and B
• $n(A \cap C)$ = number of persons who used both products A and C
• $n(B \cap C)$ = number of persons who used both products B and C
• $n(A \cap B \cap C)$ = number of persons who used all three products A, B, and C

Given Data:

• $n(A) = 97$
• $n(B) = 89$
• $n(C) = 137$
• $n(A \text{ only}) = 43$
• $n(B \text{ only}) = 10$
• $n(C \text{ only}) = 39$
• $n(B \cap C) = 62$

a) Are there any persons who have used all three products A, B, and C?

We can calculate $n(A \cap B \cap C)$ by using the principle of inclusion-exclusion.

1. Finding the number of people who used only A and B:

$n(A \cap B \text{ only}) = n(A \cap B) - n(A \cap B \cap C)$

2. Finding the number of people who used only A and C:

$n(A \cap C \text{ only}) = n(A \cap C) - n(A \cap B \cap C)$

3. Finding the number of people who used only B and C:

$n(B \cap C \text{ only}) = n(B \cap C) - n(A \cap B \cap C)$

To compute $n(A \cap B \cap C)$, we need to express all used combinations:

$n(A) = n(A \text{ only}) + n(A \cap B \text{ only}) + n(A \cap C \text{ only}) + n(A \cap B \cap C)$

$n(B) = n(B \text{ only}) + n(A \cap B \text{ only}) + n(B \cap C \text{ only}) + n(A \cap B \cap C)$

$n(C) = n(C \text{ only}) + n(A \cap C \text{ only}) + n(B \cap C \text{ only}) + n(A \cap B \cap C)$

Using the total number of persons surveyed, we have:

$n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(A \cap C) + n(A \cap B \cap C)$

Substitute the known values:

$n(A \cup B \cup C) = 97 + 89 + 137 - n(A \cap B) - 62 - n(A \cap C) + n(A \cap B \cap C)$

We also know that:

$n(A \cup B \cup C) = 217 - x \quad \text{where } x \text{ is the number of people who have not used any product.}$

Using the information on the only users:

$97 - 43 = 54 \implies n(A \cap B) + n(A \cap C) - n(A \cap B \cap C) = 54$

$89 - 10 = 79 \implies n(A \cap B) + n(B \cap C) - n(A \cap B \cap C) = 79$

$137 - 39 = 98 \implies n(A \cap C) + n(B \cap C) - n(A \cap B \cap C) = 98$

Simultaneous solving we get:

$n(A \cap B \cap C) = 10$

b) How many persons have not used any of the products A, B, and C?

We use:

$x = 217 - n(A \cup B \cup C) = 217 - (97 + 89 + 137 - 54 - 79 - 98 + 10) = 217 - 202 = 15$

Thus, 15 people haven't used any of the products.

c) How many persons have used products B or C?

$n(B \cup C) = n(B) + n(C) - n(B \cap C) = 89 + 137 - 62 = 164$

Final Answers:

a) Yes, 10 people have used all three products A, B, and C.

b) 15 persons have not used any of the products A, B, and C.

c) 164 persons have used products B or C.

Would you like more details or have any questions?

---

Related Questions:

1. How many people used exactly two products?
2. What is the probability that a randomly selected person used only product A?
3. How many people used product A or B but not C?
4. How can the principle of inclusion-exclusion be applied to find the number of people who used two or more products?
5. If another product D was introduced, how would the survey data change?

Tip:

For solving problems involving sets and surveys, the principle of inclusion-exclusion is very useful to account for overlapping sets.