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) =

1 answer

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: \[ P(W, W, W, C) \approx 0.0962 \]


  1. 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