Question

Multiple-choice questions each have 6 possible answers, one of which is correct. Assume that you guess the answers to 4 such questions.

Use the multiplication rule to find the probability that the first three guesses are wrong and the fourth is correct. That is, find
, where C denotes a correct answer and W denotes a wrong answer.
(round answer to 4 decimal places)



What is the probability of getting exactly one correct answer when 4 guesses are made?
(round answer to 4 decimal places)
P(exactly one correct answer) =

Answer 1

To find the probabilities requested, we will use basic probability principles.

1. **Finding the Probability that the First Three Guesses Are Wrong and the Fourth Is Correct:**

Each question has 6 possible answers, with only 1 correct answer and 5 incorrect answers.

- The probability of guessing wrong (W) on any question is:
\[
P(W) = \frac{5}{6}
\]

- The probability of guessing correctly (C) on any question is:
\[
P(C) = \frac{1}{6}
\]

Now, we need the probability that the first three guesses are wrong, and the fourth guess is correct:
\[
P(W, W, W, C) = P(W) \times P(W) \times P(W) \times P(C)
\]
\[
P(W, W, W, C) = P(W)^3 \times P(C)
\]
Plugging in the values:
\[
P(W, W, W, C) = \left(\frac{5}{6}\right)^3 \times \frac{1}{6}
\]
Calculating that:
\[
P(W, W, W, C) = \left(\frac{5^3}{6^3}\right) \times \frac{1}{6} = \frac{125}{216} \times \frac{1}{6} = \frac{125}{1296}
\]
Calculating \( \frac{125}{1296} \):
\[
\frac{125}{1296} \approx 0.0962
\]

**Rounded to 4 decimal places**:
\[
P(W, W, W, C) \approx 0.0962
\]

---

2. **Finding the Probability of Getting Exactly One Correct Answer When 4 Guesses Are Made:**

To find this probability, we can use the binomial probability formula, where:
- \( n = 4 \) (number of trials)
- \( k = 1 \) (number of successes)
- \( p = \frac{1}{6} \) (probability of success)
- \( q = \frac{5}{6} \) (probability of failure)

The binomial probability formula is given by:
\[
P(X = k) = \binom{n}{k} p^k q^{n-k}
\]
For our situation:
\[
P(X = 1) = \binom{4}{1} \left(\frac{1}{6}\right)^1 \left(\frac{5}{6}\right)^{4-1}
\]
Calculating \( \binom{4}{1} = 4 \):
\[
P(X = 1) = 4 \times \frac{1}{6} \times \left(\frac{5}{6}\right)^3
\]
Calculating \( \left(\frac{5}{6}\right)^3 \):
\[
\left(\frac{5}{6}\right)^3 = \frac{125}{216}
\]
Thus,
\[
P(X = 1) = 4 \times \frac{1}{6} \times \frac{125}{216} = \frac{500}{1296}
\]
Calculating \( \frac{500}{1296} \):
\[
\frac{500}{1296} \approx 0.3862
\]

**Rounded to 4 decimal places**:
\[
P(\text{exactly one correct answer}) \approx 0.3862
\]

### Summary of Results:
1. Probability that the first three guesses are wrong and the fourth is correct: **0.0962**
2. Probability of getting exactly one correct answer: **0.3862**