The question is asking for the probability of getting exactly 4 questions correct out of 12 on a multiple-choice test where each question has 5 possible answers (only 1 is correct), assuming you are guessing. This is a binomial probability problem.

Step-by-step Solution:

1. Define the Variables:

   • Total number of questions: $n = 12$
   • Number of correct answers desired: $k = 4$
   • Probability of getting a single question correct: $p = \frac{1}{5} = 0.2$
   • Probability of getting a single question wrong: $q = 1 - p = 0.8$

2. Binomial Probability Formula: The probability of getting exactly $k$ successes in $n$ trials is given by: $P(X = k) = \binom{n}{k} \cdot p^k \cdot q^{n-k}$ where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ is the number of ways to choose $k$ successes from $n$ trials.

3. Substitute Values: $P(X = 4) = \binom{12}{4} \cdot (0.2)^4 \cdot (0.8)^8$

4. Compute Each Term:

   • Compute $\binom{12}{4}$: $\binom{12}{4} = \frac{12!}{4!(12-4)!} = \frac{12 \cdot 11 \cdot 10 \cdot 9}{4 \cdot 3 \cdot 2 \cdot 1} = 495$
   • Compute $(0.2)^4$: $(0.2)^4 = 0.0016$
   • Compute $(0.8)^8$: $(0.8)^8 = 0.16777216$

5. Combine Everything: $P(X = 4) = 495 \cdot 0.0016 \cdot 0.16777216$ $P(X = 4) \approx 0.1329$

Final Answer:

The probability of getting exactly 4 questions correct is approximately 0.1329.

Would you like further details or have any questions?

Related Questions:

1. What is the probability of getting exactly 6 questions correct in this scenario?
2. How would the probability change if there were only 10 questions instead of 12?
3. What is the probability of getting no questions correct (0 correct answers)?
4. How would this problem differ if each question had only 4 possible answers?
5. Can you explain why binomial probabilities involve combinations?

Tip:

When solving binomial probability problems, always ensure you carefully identify $n$, $k$, $p$, and $q$ before proceeding to calculations.