For one toss of a certain coin, the probability that the outcome is heads is 0.6. If this coin is...

1. Translate the problem requirements: "At least 4 heads in 5 tosses" means we need either exactly 4 heads OR exactly 5 heads. We need to find the probability of this combined outcome.
2. Identify the two favorable scenarios: Break down "at least 4" into its component cases - exactly 4 heads (with 1 tail) and exactly 5 heads (with 0 tails).
3. Calculate probability for exactly 5 heads: This is straightforward since there's only one way to get all heads - multiply the probability of heads for each toss.
4. Calculate probability for exactly 4 heads: Determine how many ways we can arrange 4 heads and 1 tail, then multiply by the probability of each specific arrangement.
5. Apply the addition principle: Since these scenarios are mutually exclusive, add their probabilities to get the final answer.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what "at least 4 heads in 5 tosses" actually means in plain English.

When we say "at least 4," we mean "4 or more." Since we're only tossing the coin 5 times, "4 or more" can only be:

• Exactly 4 heads (and 1 tail), OR
• Exactly 5 heads (and 0 tails)

There's no way to get 6 or 7 heads when we only toss 5 times, so these are our only two possibilities.

Since these two outcomes cannot happen at the same time (they are mutually exclusive), we'll need to find the probability of each one separately, then add them together.

Process Skill: TRANSLATE - Converting "at least 4" into specific countable outcomes

2. Identify the two favorable scenarios

Now let's be very clear about our two scenarios:

Scenario 1: Exactly 4 heads and 1 tail
This means in our sequence of 5 tosses, we get heads 4 times and tails 1 time. The tail could occur on any of the 5 tosses.

Scenario 2: Exactly 5 heads and 0 tails
This means all 5 tosses result in heads.

We need to calculate the probability of each scenario happening, then add them together since we want either one OR the other to occur.

3. Calculate probability for exactly 5 heads

Let's start with the simpler case: getting heads on all 5 tosses.

For this to happen:

• First toss must be heads: probability = 0.6
• Second toss must be heads: probability = 0.6
• Third toss must be heads: probability = 0.6
• Fourth toss must be heads: probability = 0.6
• Fifth toss must be heads: probability = 0.6

Since each toss is independent, we multiply these probabilities together:
P(exactly 5 heads) = \(0.6 \times 0.6 \times 0.6 \times 0.6 \times 0.6 = (0.6)^5\)

4. Calculate probability for exactly 4 heads

This is more complex because the single tail can occur in different positions.

First, let's think about how many different ways we can arrange 4 heads and 1 tail in 5 tosses:

• The tail could be on toss 1: THHH
• The tail could be on toss 2: HTHH
• The tail could be on toss 3: HHTH
• The tail could be on toss 4: HHHT
• The tail could be on toss 5: HHHT

That's 5 different arrangements.

Now, what's the probability of any one specific arrangement (like HTHH)?
Probability = \(0.6 \times 0.4 \times 0.6 \times 0.6 \times 0.6 = (0.6)^4 \times (0.4)^1\)

Since there are 5 different ways this can happen, and each has the same probability:
P(exactly 4 heads) = \(5 \times (0.6)^4 \times (0.4)\)

Process Skill: CONSIDER ALL CASES - Systematically counting all possible arrangements

5. Apply the addition principle

Since we want "exactly 4 heads OR exactly 5 heads," and these events cannot happen simultaneously, we add their probabilities:

P(at least 4 heads) = P(exactly 4 heads) + P(exactly 5 heads)
P(at least 4 heads) = \(5(0.6)^4(0.4) + (0.6)^5\)

Looking at our answer choices, this matches option E exactly.

Final Answer

The probability of getting heads at least 4 times in 5 tosses is:
\(5(0.6)^4(0.4) + (0.6)^5\)

This corresponds to Answer Choice E.

To verify: The first term represents the 5 ways to get exactly 4 heads, and the second term represents the 1 way to get exactly 5 heads.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting "at least 4" as "exactly 4"
Students often confuse "at least 4" with "exactly 4" and only calculate the probability for getting exactly 4 heads, missing the scenario where all 5 tosses result in heads. This leads them to incomplete solutions that don't account for all favorable outcomes.

2. Attempting to use complement probability incorrectly
Some students try to solve this by calculating 1 - P(0, 1, 2, or 3 heads), which is mathematically correct but much more complex and error-prone. They may start this approach but get overwhelmed by the multiple calculations required, leading to mistakes or incomplete solutions.

3. Not recognizing this as a binomial probability problem
Students might approach this as a simple multiplication problem without understanding that they need to account for different arrangements of heads and tails. They may try to solve it using basic probability rules without considering the combinatorial aspect.

Errors while executing the approach

1. Incorrect counting of arrangements for exactly 4 heads
When calculating the number of ways to arrange 4 heads and 1 tail in 5 tosses, students might miscount and use 4 instead of 5, or use other incorrect values. This leads them to expressions like \(4(0.6)^4(0.4)\) instead of the correct \(5(0.6)^4(0.4)\).

2. Using wrong probability values for tails
Some students might use 0.6 for both heads and tails, forgetting that P(tails) = \(1 - P(heads) = 1 - 0.6 = 0.4\). This error would lead to expressions with only 0.6 values throughout their calculations.

3. Incorrect probability multiplication for specific sequences
Students might make errors in calculating the probability of a specific sequence like HHHHT, perhaps forgetting to include the tail probability or using the wrong powers for the head probability.

Errors while selecting the answer

1. Selecting only the "exactly 5 heads" component
After correctly identifying both scenarios (exactly 4 and exactly 5 heads), students might forget to add them together and select choice A, which represents only \((0.6)^5\) - the probability of getting exactly 5 heads.

2. Choosing the wrong coefficient for the "exactly 4 heads" term
Even if students understand they need to add both probabilities, they might select choice D with the coefficient 4 instead of choice E with the coefficient 5, due to their miscounting of arrangements in the execution phase.