Let’s analyze the question.

The problem asks for the probability of getting at least one question wrong on a 6-question multiple-choice test, where each question has 2 answers. Here's how we calculate:

Step-by-Step Solution:

1. Probability of getting a single question correct:

   • Since each question has 2 answers, one of which is correct, the probability of getting a question correct is: $P(\text{Correct}) = \frac{1}{2}$

2. Probability of getting a single question wrong:

   • The probability of getting a question wrong is: $P(\text{Wrong}) = 1 - P(\text{Correct}) = 1 - \frac{1}{2} = \frac{1}{2}$

3. Probability of getting all 6 questions correct:

   • The probability of answering all 6 questions correctly is: $P(\text{All Correct}) = \left(\frac{1}{2}\right)^6 = \frac{1}{64}$

4. Probability of getting at least one question wrong:

   • The complement of getting all questions correct is getting at least one question wrong: $P(\text{At least one wrong}) = 1 - P(\text{All Correct})$ Substituting $P(\text{All Correct}) = \frac{1}{64}$: $P(\text{At least one wrong}) = 1 - \frac{1}{64} = \frac{64}{64} - \frac{1}{64} = \frac{63}{64}$

Final Answer:

The probability of getting at least one question wrong is: $\boxed{\frac{63}{64}}$

Would you like me to break this further or clarify any part?

Related Questions:

1. What is the probability of getting exactly one question wrong in this scenario?
2. How does the probability change if the number of questions increases to 10?
3. What if each question had 4 answer choices instead of 2?
4. Can we calculate the expected number of questions answered incorrectly?
5. How does this problem relate to the concept of complementary probability?

Tip:

Always think about the complement when calculating probabilities for "at least one" scenarios; it often simplifies the computation!