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Chapter 3: Problem 19

Let an unbiased die be cast at random seven independent times. Compute the conditional probability that each side appears at least once given that side 1 appears exactly twice.

Short Answer

Expert verified
The conditional probability that each side appears at least once given that side 1 appears exactly twice is approximately 0.022.

Step by step solution

01

Determine the event A and B

We define two events: \n A: each side appears at least once \n B: side one appears exactly twice. \n We are interested in \( P(A|B) \), the probability of event A occurring given that B has already occurred.
02

Use the formula of conditional probability

We know that the formula of conditional probability is \( P(A|B) = \frac{P(A \cap B)}{P(B)} \). This formula will give us the probability of A given B.
03

Calculate the probability of event B

Event B (side one appears exactly twice) can happen in \( \binom{7}{2} \) ways, as we choose where the two ones will appear in our seven trials. So, \( P(B) = \frac{\binom{7}{2} * (\frac{1}{6})^2 * (\frac{5}{6})^5}{1} = 0.323 \) approximately. We divide by 1 because there's a total of one possible outcome set.
04

Calculate the probability of event A and B happening together

Event A and B happening together means that one appears twice, and each of other five numbers appears at least once in the remaining five throws. This can happen in \( \binom{7}{2} \) ways to choose where the two ones will appear and \( 5! \) ways to arrange the other five numbers in remaining five throws. So, \( P(A \cap B) = \frac{\binom{7}{2} * 5! * (\frac{1}{6})^7}{1} = 0.007 \) approximately.
05

Use the formula of conditional probability to get the result

Substitute \( P(A \cap B) \) and \( P(B) \) into \( P(A|B) = \frac{P(A \cap B)}{P(B)} \). So, \( P(A|B) = \frac{0.007}{0.323} = 0.022 \) approximately.

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