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 in 3 coin flips, where each flip has a 50% chance of heads or tails
2. Enumerate all possible outcomes: List out every possible sequence of 3 coin flips to establish the total sample space
3. Identify favorable outcomes: From all possible sequences, count only those that contain exactly 2 heads and 1 tail
4. Calculate probability using basic ratio: Apply the fundamental probability formula of favorable outcomes divided by total outcomes

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding exactly what we're looking for. We have a fair coin that we're flipping 3 times in a row. Each time we flip it, there's a 50% chance it lands heads up and a 50% chance it lands tails up. We want to find the probability that out of these 3 flips, exactly 2 will be heads and 1 will be tails.

Think of it like this: if you were to flip a coin 3 times right now, what are the chances that you'd get heads exactly twice? Not once, not three times, but exactly twice.

Process Skill: TRANSLATE - Converting the problem language into a clear mathematical question we can work with

2. Enumerate all possible outcomes

To solve this, let's think about every possible way 3 coin flips could turn out. Each flip can be either Heads (H) or Tails (T), so let's list all possibilities:

Flip 1, Flip 2, Flip 3:
• H, H, H (all heads)
• H, H, T (two heads, one tail)
• H, T, H (two heads, one tail)
• H, T, T (one head, two tails)
• T, H, H (two heads, one tail)
• T, H, T (one head, two tails)
• T, T, H (one head, two tails)
• T, T, T (all tails)

So there are 8 total possible outcomes when flipping a coin 3 times. Each of these outcomes is equally likely to happen.

Process Skill: CONSIDER ALL CASES - Systematically listing every possible sequence to ensure we don't miss any

3. Identify favorable outcomes

Now let's look at our list and circle the outcomes that give us exactly 2 heads:

• H, H, H ❌ (this has 3 heads, not 2)
• H, H, T ✓ (exactly 2 heads!)
• H, T, H ✓ (exactly 2 heads!)
• H, T, T ❌ (this has only 1 head)
• T, H, H ✓ (exactly 2 heads!)
• T, H, T ❌ (this has only 1 head)
• T, T, H ❌ (this has only 1 head)
• T, T, T ❌ (this has 0 heads)

So we have exactly 3 favorable outcomes out of 8 total possible outcomes. The favorable outcomes are: HHT, HTH, and THH.

4. Calculate probability using basic ratio

Probability is simply the number of ways something can happen divided by the total number of possible outcomes.

In plain English: Out of 8 equally likely ways to flip 3 coins, exactly 3 of those ways give us 2 heads.

Probability = Number of favorable outcomes ÷ Total number of outcomes
Probability = \(3 \div 8 = 0.375\)

Let's verify this matches our answer choices: 0.375 corresponds to choice C.

Final Answer

The probability that a fair coin will land heads up exactly twice in 3 consecutive flips is 0.375.

This matches answer choice C. 0.375, which we can verify by noting that \(\frac{3}{8} = 0.375\).

Common Faltering Points

Errors while devising the approach

1. Misinterpreting "exactly twice" as "at least twice"
Students may confuse the requirement of getting heads "exactly twice" with "at least twice" or "at most twice." This leads them to include scenarios with 3 heads (HHH) in their favorable outcomes, fundamentally changing the problem they're solving.

2. Attempting to use formulas without understanding
Some students might immediately jump to using the binomial probability formula \(\mathrm{C(n,r)} \times \mathrm{p}^r \times (1-\mathrm{p})^{(n-r)}\) without fully understanding what they're calculating. While this approach would work, rushing to formulas without conceptual understanding often leads to substitution errors.

Errors while executing the approach

1. Incomplete enumeration of outcomes
When listing all possible sequences of 3 coin flips, students commonly miss one or more outcomes. For example, they might list HHT and HTH but forget THH, leading to an incorrect count of favorable outcomes (2 instead of 3).

2. Miscounting favorable outcomes
Even with a complete list, students may incorrectly identify which outcomes have exactly 2 heads. They might accidentally include outcomes like HTT (which has only 1 head) or exclude valid outcomes like THH, leading to wrong numerators in their probability calculation.

3. Calculating total outcomes incorrectly
Students might miscalculate the total number of possible outcomes for 3 coin flips. Instead of \(2^3 = 8\), they might think there are only 6 outcomes or confuse this with the number of ways to arrange 2 heads and 1 tail.

Errors while selecting the answer

1. Converting fraction to decimal incorrectly
After correctly finding that the probability is \(\frac{3}{8}\), students may make arithmetic errors when converting to decimal form. They might calculate \(3 \div 8\) incorrectly, getting 0.25 or 0.50 instead of 0.375, leading them to select answer choice A or D.