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)
0.0962 is not the answer
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)
0.0962 is not the answer
Answers
GPT-4o mini
To find the probability that the first three guesses are wrong and the fourth is correct, we can use the multiplication rule of probability.
1. The probability of guessing a wrong answer (W) for a single question is:
\[
P(W) = \frac{5}{6}
\]
(since there are 5 wrong answers out of 6 possible answers).
2. The probability of guessing a correct answer (C) for a single question is:
\[
P(C) = \frac{1}{6}
\]
3. We need to find the probability of the sequence: W, W, W, C. According to the multiplication rule:
\[
P(W) \times P(W) \times P(W) \times P(C) = P(W)^3 \times P(C)
\]
4. Now, substituting the probabilities:
\[
P(W)^3 = \left(\frac{5}{6}\right)^3 = \frac{125}{216}
\]
\[
P(C) = \frac{1}{6}
\]
Thus:
\[
P(W, W, W, C) = \left(\frac{5}{6}\right)^3 \times \left(\frac{1}{6}\right) = \frac{125}{216} \times \frac{1}{6} = \frac{125}{1296}
\]
5. Now, we need to calculate this value:
\[
\frac{125}{1296} \approx 0.0964
\]
Rounding this to four decimal places gives:
\[
\boxed{0.0964}
\]
1. The probability of guessing a wrong answer (W) for a single question is:
\[
P(W) = \frac{5}{6}
\]
(since there are 5 wrong answers out of 6 possible answers).
2. The probability of guessing a correct answer (C) for a single question is:
\[
P(C) = \frac{1}{6}
\]
3. We need to find the probability of the sequence: W, W, W, C. According to the multiplication rule:
\[
P(W) \times P(W) \times P(W) \times P(C) = P(W)^3 \times P(C)
\]
4. Now, substituting the probabilities:
\[
P(W)^3 = \left(\frac{5}{6}\right)^3 = \frac{125}{216}
\]
\[
P(C) = \frac{1}{6}
\]
Thus:
\[
P(W, W, W, C) = \left(\frac{5}{6}\right)^3 \times \left(\frac{1}{6}\right) = \frac{125}{216} \times \frac{1}{6} = \frac{125}{1296}
\]
5. Now, we need to calculate this value:
\[
\frac{125}{1296} \approx 0.0964
\]
Rounding this to four decimal places gives:
\[
\boxed{0.0964}
\]
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