The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to...

1. Translate the problem requirements: We need to find the probability that out of 3 independent visitors, exactly 2 will buy candy (with 30% individual probability) and 1 will not buy candy (with 70% individual probability).
2. Identify all possible scenarios: List out the specific ways exactly 2 out of 3 visitors can buy candy, considering which visitor doesn't buy.
3. Calculate probability for each scenario: For each identified scenario, multiply the individual probabilities of visitors buying or not buying candy.
4. Sum the probabilities: Add up the probabilities of all scenarios since they are mutually exclusive events that achieve our desired outcome.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we know from the problem:

• Each visitor has a 30% chance of buying candy (which means a 70% chance of NOT buying candy)
• We have 3 visitors coming to the mall
• We want to find the probability that exactly 2 out of these 3 visitors will buy candy

This means exactly 2 visitors buy candy AND exactly 1 visitor does not buy candy.

Process Skill: TRANSLATE - Converting the word problem into clear mathematical requirements

2. Identify all possible scenarios

Let's think about this concretely. If we call our three visitors A, B, and C, then exactly 2 buying candy means:

• Scenario 1: Visitor A buys, Visitor B buys, Visitor C does NOT buy
• Scenario 2: Visitor A buys, Visitor B does NOT buy, Visitor C buys
• Scenario 3: Visitor A does NOT buy, Visitor B buys, Visitor C buys

These are the only three ways to have exactly 2 out of 3 visitors buy candy. Notice that in each scenario, exactly one visitor doesn't buy candy, but it's a different visitor each time.

Process Skill: CONSIDER ALL CASES - Systematically listing all possible ways to achieve our desired outcome

3. Calculate probability for each scenario

Now let's calculate the probability for each scenario. Remember:

• Probability of buying candy = 0.3
• Probability of NOT buying candy = 0.7

Scenario 1: A buys × B buys × C doesn't buy
Probability = \(0.3 \times 0.3 \times 0.7 = 0.063\)

Scenario 2: A buys × B doesn't buy × C buys
Probability = \(0.3 \times 0.7 \times 0.3 = 0.063\)

Scenario 3: A doesn't buy × B buys × C buys
Probability = \(0.7 \times 0.3 \times 0.3 = 0.063\)

Notice that all three scenarios have the same probability! This makes sense because each scenario involves exactly two people buying \(0.3 \times 0.3\) and exactly one person not buying (0.7).

4. Sum the probabilities

Since these three scenarios are the only ways to get exactly 2 visitors buying candy, and they cannot happen at the same time (they are mutually exclusive), we add their probabilities together:

Total probability = \(0.063 + 0.063 + 0.063 = 0.189\)

We can also think of this as: \(3 \times 0.063 = 0.189\)

The factor of 3 comes from the fact that there are 3 different ways to choose which 2 visitors out of 3 will buy candy.

Final Answer

The probability that exactly two visitors will buy a pack of candy is 0.189.

Looking at our answer choices, this matches choice C) 0.189.

Common Faltering Points

Errors while devising the approach

Faltering Point 1: Misinterpreting "exactly 2" as "at least 2"
Students may calculate the probability for 2 OR MORE visitors buying candy instead of EXACTLY 2. This would lead them to calculate \(\mathrm{P(exactly\ 2)} + \mathrm{P(exactly\ 3)}\), giving an incorrect higher probability. The word "exactly" is crucial and means we only want the scenario where 2 visitors buy candy, not 2 or 3.

Faltering Point 2: Not recognizing this as a binomial probability problem
Some students might try to use simple multiplication \(0.3 \times 0.3\) without considering that there are multiple ways to select which 2 out of 3 visitors will buy candy. They miss that this is a binomial probability situation requiring the combination formula or systematic enumeration of scenarios.

Errors while executing the approach

Faltering Point 1: Using wrong probability values
Students may incorrectly use 0.7 (probability of NOT buying) instead of 0.3 (probability of buying) when calculating scenarios, or vice versa. For example, calculating Scenario 1 as \(0.7 \times 0.7 \times 0.3\) instead of \(0.3 \times 0.3 \times 0.7\).

Faltering Point 2: Missing one or more scenarios
When listing all possible ways exactly 2 visitors can buy candy, students might only identify 1 or 2 scenarios instead of all 3. For instance, they might only consider "first two buy, third doesn't" and forget about the other arrangements.

Faltering Point 3: Arithmetic calculation errors
Students may make computational mistakes when calculating \(0.3 \times 0.3 \times 0.7 = 0.063\), or when adding the three scenario probabilities: \(0.063 + 0.063 + 0.063 = 0.189\). These decimal multiplications are prone to calculation errors.

Errors while selecting the answer

Faltering Point 1: Selecting the probability of a single scenario instead of the total
Students might calculate one scenario correctly (getting 0.063) but then select answer choice D) 0.063 instead of recognizing they need to add all three scenarios together to get the final answer of 0.189.