If a coin has an equal probability of landing heads up or tails up each time it is flipped, what...

1. Translate the problem requirements: We need to find the probability of getting exactly 2 heads (and therefore 1 tail) when flipping a fair coin 3 times in a row. Each flip has probability \(\frac{1}{2}\) for heads and \(\frac{1}{2}\) for tails.
2. Identify all possible favorable outcomes: List out all the different ways we can arrange exactly 2 heads and 1 tail in 3 consecutive flips.
3. Calculate probability for each favorable outcome: Since each outcome involves specific results for 3 independent flips, multiply the individual probabilities.
4. Sum up all favorable probabilities: Add the probabilities of all favorable outcomes to get the total probability.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what this problem is asking us to find. We have a fair coin, which means it has an equal chance of landing heads or tails on each flip. When we flip it once, there's a 50% chance (or \(\frac{1}{2}\) probability) it lands heads, and a 50% chance it lands tails.

We're flipping this coin 3 times in a row, and we want to know: what's the probability that we get exactly 2 heads and 1 tail in those 3 flips?

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

2. Identify all possible favorable outcomes

Let's think about this step by step. If we flip a coin 3 times and want exactly 2 heads, we need to figure out all the different ways this can happen.

Let's use H for heads and T for tails. The different arrangements are:

• First scenario: H-H-T (heads on first flip, heads on second flip, tails on third flip)
• Second scenario: H-T-H (heads on first flip, tails on second flip, heads on third flip)
• Third scenario: T-H-H (tails on first flip, heads on second flip, heads on third flip)

These are the only three ways to get exactly 2 heads in 3 flips. Notice that in each case, we have exactly 2 H's and 1 T.

Process Skill: CONSIDER ALL CASES - Systematically listing all favorable outcomes

3. Calculate probability for each favorable outcome

Now let's figure out the probability of each scenario happening. Since each flip is independent, we multiply the probabilities of individual flips.

For scenario H-H-T:

• Probability of heads on first flip: \(\frac{1}{2}\)
• Probability of heads on second flip: \(\frac{1}{2}\)
• Probability of tails on third flip: \(\frac{1}{2}\)
• Combined probability: \(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}\)

For scenario H-T-H:

• Probability: \(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}\)

For scenario T-H-H:

• Probability: \(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}\)

Notice that each favorable scenario has the same probability: \(\frac{1}{8}\). This makes sense because each scenario involves exactly 2 heads and 1 tail, just in different orders.

4. Sum up all favorable probabilities

To find the total probability of getting exactly 2 heads in 3 flips, we add up the probabilities of all our favorable scenarios:

Total probability = \(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{3}{8}\)

Let's convert this to a decimal: \(\frac{3}{8} = 0.375\)

Looking at our answer choices, 0.375 corresponds to choice C.

Final Answer

The probability that the coin will land heads up exactly twice in 3 consecutive flips is \(\frac{3}{8} = 0.375\).

The correct answer is C.

Common Faltering Points

Errors while devising the approach

Faltering Point 1: Misunderstanding "exactly twice"
Students may confuse "exactly twice" with "at least twice" and include scenarios where all 3 flips are heads. The question specifically asks for exactly 2 heads (not 2 or more), meaning we must have exactly 1 tail as well.

Faltering Point 2: Thinking order doesn't matter
Some students might think there's only one way to get 2 heads in 3 flips, not realizing that H-H-T, H-T-H, and T-H-H are three distinct outcomes. They might incorrectly assume the probability is simply \(\left(\frac{1}{2}\right)^3\) instead of considering all possible arrangements.

Faltering Point 3: Confusing this with a combinations problem
Students might try to use combination formulas like \(\mathrm{C}(3,2)\) without understanding that they still need to calculate the actual probability of each arrangement occurring.

Errors while executing the approach

Faltering Point 1: Arithmetic errors in fraction operations
When calculating \(\left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) = \frac{1}{8}\) for each scenario, students might make basic multiplication errors or struggle with adding fractions: \(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{3}{8}\).

Faltering Point 2: Converting fractions to decimals incorrectly
Students may make errors when converting \(\frac{3}{8}\) to 0.375, potentially getting 0.38 or 0.35, leading them to select an incorrect answer choice.

Faltering Point 3: Missing one or more favorable scenarios
Students might correctly identify that H-H-T has probability \(\frac{1}{8}\) but forget to list all three arrangements (H-H-T, H-T-H, T-H-H), leading to an incomplete calculation and selecting 0.125 instead of 0.375.

Errors while selecting the answer

Faltering Point 1: Selecting the probability of a single scenario
After correctly calculating that each favorable scenario has probability \(\frac{1}{8} = 0.125\), students might mistakenly select answer choice A (0.125) instead of recognizing they need to add up all three scenarios to get 0.375.

Faltering Point 2: Confusing \(\frac{3}{8}\) with \(\frac{3}{6}\) or other fractions
Students might misread their final calculation and think they got \(\frac{3}{6} = 0.5\) (answer choice D) instead of the correct \(\frac{3}{8} = 0.375\).