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Chapter 3: Problem 44

Each of 2 cabinets identical in appearance has 2 drawers. Cabinet \(A\) contains a silver coin in each drawer, and cabinet \(B\) contains a silver coin in one of its drawers and a gold coin in the other. A cabinet is randomly selected, one of its drawers is opened, and a silver coin is found. What is the probability that there is a silver coin in the other drawer?

Short Answer

Expert verified
The probability of having a silver coin in the other drawer given that we found a silver coin is \(\frac{2}{3}\).

Step by step solution

01

Understand Bayes' theorem for conditional probability

Bayes' theorem relates the probabilities of events A and B and can be written as follows for the conditional probability P(A|B): \(P(A|B) = \frac{P(B|A) * P(A)}{P(B)}\) In our problem, we want to find the probability that Cabinet A was chosen, given that a silver coin was found. Thus, we can define the events A and B as follows: - Event A: Cabinet A was chosen - Event B: A silver coin was found Our goal is to find P(A|B), the probability of choosing Cabinet A given that we found a silver coin.
02

Calculate the probabilities P(A), P(B|A), and P(B)

To apply Bayes' theorem, we first need to calculate the probabilities P(A), P(B|A), and P(B). - P(A): Since both cabinets are identical and one of them is randomly chosen, the probability of choosing Cabinet A is \(\frac{1}{2}\). - P(B|A): If we chose Cabinet A (which has two silver coins), the probability of finding a silver coin is 1. - P(B): To find the probability of finding a silver coin, we can consider both ways this can take place: 1) finding a silver coin in Cabinet A, or 2) finding a silver coin in Cabinet B. Both cabinets have a \(\frac{1}{2}\) chance of being chosen. So the probability of finding a silver coin in Cabinet A is \(\frac{1}{2} * 1 = \frac{1}{2}\) and the probability of finding a silver coin in Cabinet B is \(\frac{1}{2} * \frac{1}{2} = \frac{1}{4}\). Then, the total probability of finding a silver coin, P(B), is the sum of these probabilities: \(\frac{1}{2} + \frac{1}{4} = \frac{3}{4}\). Now we have the necessary values to apply Bayes' theorem.
03

Apply Bayes' theorem to find P(A|B)

Now we can apply Bayes' theorem to find the conditional probability P(A|B): \(P(A|B) = \frac{P(B|A) * P(A)}{P(B)} = \frac{1 * \frac{1}{2}}{\frac{3}{4}} = \frac{\frac{1}{2}}{\frac{3}{4}} = \frac{1}{2} * \frac{4}{3} = \frac{2}{3}\) The probability of having a silver coin in the other drawer given that we found a silver coin is \(\frac{2}{3}\).

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Most popular questions from this chapter

Prove or give counterexamples to the following statements: (a) If \(E\) is independent of \(F\) and \(E\) is independent of \(G\), then \(E\) is independent of \(F \cup G\). (b) If \(E\) is independent of \(F\), and \(E\) is independent of \(G\), and \(F G=\varnothing\), then \(E\) is independent of \(F \cup G\). (c) If \(E\) is independent of \(F\), and \(F\) is independent of \(G\), and \(E\) is independent of \(F G\), then \(G\) is independent of \(E F\).

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