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Chapter 3: Problem 48

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 that there is a silver coin in the other drawer, given that a silver coin is found in one drawer, is \( \frac{2}{3} \).

Step by step solution

01

Identify the sample space and events

Let S be the sample space, A be the event that cabinet A is chosen containing two silver coins, and B be the event that cabinet B is chosen containing one silver coin and one gold coin. The sample space consists of the following possible outcomes: two silver coins in cabinet A, or one silver and one gold coin in cabinet B.
02

Calculate the probability of each event

Both cabinets have an equal chance of being chosen, so the probability of selecting cabinet A (P(A)) and cabinet B (P(B)) is 1/2 for each.
03

Calculate the conditional probabilities

We need the conditional probability of finding a silver coin in the other drawer given that we found a silver coin. This can be written as P(Silver in other drawer | Silver in one drawer). Using the conditional probability formula, this is equal to P(Silver in other drawer and Silver in one drawer) / P(Silver in one drawer).
04

Calculate the numerator

The numerator is the probability of getting a silver coin in both drawers. Since cabinet A has two silver coins, the probability is P(A) * 1 = 1/2 because there is a 1/2 chance of selecting cabinet A, and the probability of getting a silver coin is 1.
05

Calculate the denominator

The denominator is the probability of finding a silver coin in one drawer. Since cabinet A has two silver coins and cabinet B has one silver coin and one gold coin, the probability is P(A) * 1 + P(B) * 1/2 = (1/2) * 1 + (1/2) * (1/2) = 1/2 + 1/4 = 3/4. This is because there is a 1/2 chance of selecting cabinet A, with a 100% chance of getting a silver coin, and a 1/2 chance of selecting cabinet B with a 50% chance of getting a silver coin.
06

Calculate the conditional probability

Now we can find the conditional probability P(Silver in other drawer | Silver in one drawer) = P(Silver in other drawer and Silver in one drawer) / P(Silver in one drawer) = (1/2) / (3/4) = (1/2) * (4/3) = 2/3. So, the probability that there is a silver coin in the other drawer is \( \frac{2}{3} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Space
Understanding the concept of sample space is foundational to grasping probability. The sample space, often denoted as S, is the set of all possible outcomes in a probability experiment. Imagine it as a comprehensive list that includes every conceivable result that could occur.

For instance, if you're flipping a fair coin, there are two possible outcomes: heads or tails. Thus, the sample space for this experiment is 'heads, tails'. In the exercise provided, we have two cabinets, each with two possibilities: both drawers having silver coins, or one drawer having a silver coin and the other a gold coin. This gives us our sample space for the experiment, which is essential for calculating probabilities.
Event Probability
Event probability refers to the measure of the likelihood of a specific outcome or set of outcomes occurring within the sample space. To calculate this, we divide the number of ways an event can happen by the total number of possible outcomes. Probabilities always range from 0 (impossible event) to 1 (certain event).

For example, in the drawer scenario, the probability of selecting any given cabinet is 50%, since there are two cabinets and no information is given to differentiate their likelihood of selection. When we found a silver coin in one drawer, it affected the likelihood of what is found in the other drawer. The event probability is a dynamic figure that is influenced by the observed outcomes, a concept that is often assessed through conditional probability, which leads us to Bayes' theorem.
Bayes' Theorem
Bayes' theorem is a powerful formula used to compute the probability of an event based on prior knowledge of conditions that might be related to the event. It鈥檚 especially useful when dealing with conditional probabilities, which are the probabilities of an event occurring given that another event has already occurred.

To apply Bayes' theorem, we require the probabilities of each event and how they intersect with each other. In our cabinet and drawer exercise, we want to determine the probability that the second coin is silver after observing the first coin is silver. This conditional probability takes into account our prior knowledge of the possible configurations of the cabinets. Bayes' theorem helps us update our belief about the likelihood of an event (finding another silver coin) based on the new evidence (first coin is silver), which is exactly what we did to arrive at the solution of the exercise.

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Most popular questions from this chapter

Two cards are randomly chosen without replacement from an ordinary deck of 52 cards. Let \(B\) be the event that both cards are aces, let \(A_{s}\) be the event that the ace of spades is chosen, and let \(A\) be the event that at least one ace is chosen. Find (a) \(P\left(B | A_{s}\right)\) (b) \(P(B | A)\)

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Suppose that an insurance company classifies people into one of three classes: good risks, average risks, and bad risks. The company's records indicate that the probabilities that good-, average-, and bad-risk persons will be involved in an accident over a 1 -year span are, respectively, .05, .15 and \(30 .\) If 20 percent of the population is a good risk, 50 percent an average risk, and 30 percent a bad risk, what proportion of people have accidents in a fixed year? If policyholder \(A\) had no accidents in \(2012,\) what is the probability that he or she is a good risk? is an average risk?

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Suppose that each child born to a couple is equally likely to be a boy or a girl, independently of the sex distribution of the other children in the family. For a couple having 5 children, compute the probabilities of the following events: (a) All children are of the same sex. (b) The 3 eldest are boys and the others girls. (c) Exactly 3 are boys. (d) The 2 oldest are girls. (e) There is at least 1 girl.
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