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a man is known to speak the lies 1 out of 3 times. he throws a die and reports that it is a six. find the probability that is actually a six.

User Maks K
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Final answer:

To find the probability that the die is actually a six when the man reports a six, we can use Bayes' theorem.

By calculating the individual probabilities and plugging them into the formula, we find that the probability is 3/7.

Step-by-step explanation:

Let's denote the event of the man reporting a six as event A, and the event of the actual roll resulting in a six as event B.

We know that the probability of speaking a lie is 1/3 and the probability of rolling a six on a fair die is 1/6.

We can use Bayes' theorem to calculate the probability:

P(B|A) = (P(A|B) * P(B)) / P(A)

P(A|B) is the probability of reporting a six given that the actual roll was a six, which is 1.

P(B) is the probability of the actual roll resulting in a six, which is 1/6.

P(A) is the probability of reporting a six, which can be calculated as:

P(A) = P(A|B) * P(B) + P(A|B') * P(B')

= 1 * 1/6 + 1/3 * 5/6

= 7/18.

Plugging these values into the Bayes' theorem formula:

P(B|A) = (1 * 1/6) / (7/18)

= 3/7.

Therefore, the probability that the actual roll is a six given that the man reports a six is 3/7.

User EvilDuck
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