A contest will consist of n questions, each of which is to be answered with "True" or "False". Anyone who...

1. Translate the problem requirements: We need the smallest number of True/False questions where randomly guessing all answers correctly has less than \(\frac{1}{1000}\) probability of success
2. Set up the probability framework: Each question has probability \(\frac{1}{2}\) of being answered correctly by random guessing, so n questions have probability \(\left(\frac{1}{2}\right)^n\) of all being correct
3. Establish the inequality condition: We need \(\left(\frac{1}{2}\right)^n < \frac{1}{1000}\), which means \(2^n > 1000\)
4. Test values systematically using answer choices: Calculate powers of 2 for the given options to find the first value where \(2^n\) exceeds 1000

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what this problem is asking in plain English. We have a contest with n True/False questions. To win, someone must get ALL n questions right. We want to find the smallest number of questions where a person randomly guessing has less than a 1 in 1000 chance of winning.

Think of it this way: if there's only 1 question, a random guesser has a 50-50 chance of winning - that's way too high! We need enough questions so that getting them all right by pure luck becomes extremely unlikely.

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

2. Set up the probability framework

Let's think through this step by step with concrete examples:

• For 1 question: Probability of getting it right = \(\frac{1}{2}\) (50% chance)
• For 2 questions: Probability of getting BOTH right = \(\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\) (25% chance)
• For 3 questions: Probability of getting ALL THREE right = \(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}\) (12.5% chance)

Do you see the pattern? Each additional question cuts the probability in half!

For n questions, the probability of getting all correct by random guessing = \(\left(\frac{1}{2}\right)^n\)

3. Establish the inequality condition

We need this probability to be less than \(\frac{1}{1000}\). In everyday terms, we want less than 1 chance in 1000 of someone randomly guessing their way to victory.

So we need: \(\left(\frac{1}{2}\right)^n < \frac{1}{1000}\)

Here's a helpful way to think about this: if \(\left(\frac{1}{2}\right)^n < \frac{1}{1000}\), then flipping this around gives us \(2^n > 1000\).

This makes sense! We need \(2^n\) (which grows very quickly) to exceed 1000.

4. Test values systematically using answer choices

Now let's test the answer choices by calculating powers of 2:

• n = 5: \(2^5 = 32\). Is \(32 > 1000\)? No, so n = 5 is not enough.
• n = 10: \(2^{10} = 1024\). Is \(1024 > 1000\)? Yes! This means \(\left(\frac{1}{2}\right)^{10} < \frac{1}{1000}\).

Let's verify: \(\left(\frac{1}{2}\right)^{10} = \frac{1}{1024}\), and \(\frac{1}{1024} \approx 0.00098\), which is indeed less than \(\frac{1}{1000} = 0.001\).

Since we want the LEAST value of n, and 10 is the smallest answer choice that works, we don't need to check the larger values (50, 100, 1000).

Process Skill: APPLY CONSTRAINTS - Testing systematically to find the minimum value that satisfies our condition

Final Answer

The least value of n for which the probability is less than \(\frac{1}{1000}\) is n = 10.

With 10 questions, a random guesser has only a \(\frac{1}{1024}\) chance (less than \(\frac{1}{1000}\)) of getting all answers correct.

The answer is B. 10.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting "all n questions correctly" as "at least some questions correctly"

Students may miss the critical word "ALL" in the problem statement. The contest requires answering ALL n questions correctly to be a winner. Some students might think that getting most questions right or a certain percentage right would qualify as winning, leading them to set up an entirely different probability calculation.

2. Setting up the inequality in the wrong direction

The problem asks for when the probability is "less than \(\frac{1}{1000}\)." Students often confuse this and set up the inequality as \(\left(\frac{1}{2}\right)^n > \frac{1}{1000}\) instead of \(\left(\frac{1}{2}\right)^n < \frac{1}{1000}\). This reversal leads to finding when the probability is greater than \(\frac{1}{1000}\), which answers the opposite question.

3. Misunderstanding what "randomly guessing" means for True/False questions

Students might incorrectly assume that random guessing gives some probability other than \(\frac{1}{2}\) for each question. They may think there are more than 2 choices or that the probability changes based on previous answers, rather than understanding that each True/False question has exactly a \(\frac{1}{2}\) probability of being correct when guessed randomly.

Errors while executing the approach

1. Computational errors with powers of 2

Students frequently make arithmetic mistakes when calculating powers of 2, especially \(2^{10} = 1024\). They might calculate \(2^{10}\) as 1000 or some other incorrect value, leading them to conclude that n = 10 doesn't satisfy the requirement when it actually does.

2. Incorrectly manipulating the inequality \(\left(\frac{1}{2}\right)^n < \frac{1}{1000}\)

When converting \(\left(\frac{1}{2}\right)^n < \frac{1}{1000}\) to \(2^n > 1000\), students often make errors with the inequality direction or the algebraic manipulation. They might forget to flip the inequality when taking reciprocals, or incorrectly rewrite the inequality as \(2^n < 1000\).

3. Confusing \(\frac{1}{1000}\) with 1000 in calculations

Students may substitute 1000 instead of \(\frac{1}{1000}\) in their calculations, or vice versa. For example, they might check if \(\left(\frac{1}{2}\right)^{10} < 1000\) instead of checking if \(\left(\frac{1}{2}\right)^{10} < \frac{1}{1000}\), leading to completely wrong conclusions about which values of n work.

Errors while selecting the answer

1. Choosing the largest value that works instead of the smallest

The question specifically asks for the "least value of n." Students who correctly identify that multiple answer choices work (like n = 10, 50, 100, 1000) might select a larger value thinking "bigger is safer" rather than the smallest value that satisfies the condition.

2. Selecting based on approximate calculations without verification

Students might do rough mental math and conclude that since \(2^{10} = 1024\) is "close to" 1000, it might not be sufficient. They may then choose n = 50 or n = 100 as a "safe" choice without properly verifying that \(\frac{1}{1024}\) is indeed less than \(\frac{1}{1000}\).

no likely faltering points