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The probability that a visitor at the mall buys a pack of candy is \(30\%\). If three visitors come to the mall today, what is the probability that exactly two visitors will buy a pack of candy?
Let's break down what we know from the problem:
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
Let's think about this concretely. If we call our three visitors A, B, and C, then exactly 2 buying candy means:
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
Now let's calculate the probability for each scenario. Remember:
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).
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.
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.
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.
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.
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.