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Consider a set of ten urns, nine of which contain three white chips and three red chips each. The tenth contains five white chips and one red chip. An urn is picked at random. Then a sample of size 3 is drawn without replacement from that urn. If all three chips drawn are white, what is the probability that the urn being sampled is the one with five white chips?

Short Answer

Expert verified
Hence, the probability that the urn being sampled is the one with five white chips given that all three chips drawn are white is approximately 0.38.

Step by step solution

01

Determine Prior Probabilities

For each of the ten urns, there is an equal probability of picking an urn, denoted by \( P(U_i) \), where \( i \) refers to the urn type. Therefore, \( P(U_1) = p(U_2) = ... = p(U_9) = 1/10 \) for the urns with three white and three red chips and \( P(U_{10}) = 1/10 \) for the urn with five white chips and one red chip.
02

Determine Likelihood Probabilities

Next, we need to determine the likelihood probability \( P(W_3 | U_i) \), which refers to the probability of drawing three white chips given that urn \( i \) was picked. For the urns with three white and three red chips, this is \( \frac{{3 \choose 3}}{{6 \choose 3}} = 0.1 \). For the urn with five white chips and one red chip, the likelihood is \( \frac{{5 \choose 3}}{{6 \choose 3}} \approx 0.56 \).
03

Determine Overall Probability

The overall probability of drawing three white chips from any urn, \( P(W_3) \), is a weighted average of the likelihood probabilities for each urn type, where the weights are the prior probabilities. So, \( P(W_3) = \frac{9}{10} \cdot 0.1 + \frac{1}{10} \cdot 0.56 = 0.146 \).
04

Apply Bayes' Theorem

Finally, we apply Bayes' theorem to determine the probability that the urn with five white chips and one red chip was picked given that three white chips were drawn, \( P(U_{10} | W_3) \). Hence, \( P(U_{10} | W_3) = \frac{P(W_3 | U_{10}) \cdot P(U_{10})}{P(W_3)} = \frac{0.56 \cdot 1/10}{0.146} \approx 0.38.

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Key Concepts

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

Conditional Probability
Conditional probability is the likelihood of an event occurring, given that another event has already occurred. This concept serves as a cornerstone in understanding complex probability scenarios where events are interlinked. Imagine you're at a sports event and want to predict the likelihood of your favorite team winning, conditional on it being a home game. This adjusted probability can differ from the probability of winning without that prior information.

In the exercise provided, conditional probability is used to calculate the chances of drawing three white chips from a specific urn, given that the urn has already been selected. This is an important concept because it allows for the refinement of probability assessments based on new evidence or conditions. Remember that the conditional probability formula is expressed as: \( P(A|B) = \frac{P(A \cap B)}{P(B)} \), where \( P(A|B) \) is the probability of event A given event B.
Likelihood Probability
Likelihood probability distinguishes itself from other probability types through its role in updating beliefs in light of new information. It quantifies how probable a specific outcome is, given a set of parameters or conditions. For example, think about diagnosing a disease; the likelihood of observing certain symptoms increases if we know the patient has the disease.

In our exercise's context, the likelihood of drawing three white chips given a specific urn (denoted as \( P(W_3 | U_i) \) in our solution) is required. These likelihood probabilities contribute significantly to Bayes' theorem, enabling it to update prior beliefs (or probabilities) in the light of observed evidence (the drawing of the three white chips in this case).
Prior Probabilities
Prior probabilities are initial assessments of the likelihood of an event before any additional evidence is considered. They are the 'priors' in Bayesian statistics, which help form a starting point for further analysis. Consider these as the base rate or default belief before looking at the specifics of a case, similar to how a doctor estimates the prevalence of a disease before examining a particular patient.

In solving this exercise, prior probabilities are the chances of selecting each urn at random (\(\ P(U_i) \)) before we draw any chips out of them. These priors are then updated by the likelihood of the observed outcome (drawing three white chips) to yield a revised probability, a process essentially captured by Bayes' theorem.
Probability Without Replacement
When discussing probability without replacement, we refer to scenarios where an item is not returned to a set after being selected. This alters the probabilities of subsequent selections since the composition of the set changes each time. For instance, removing a card from a standard deck will change the odds for drawing any remaining cards.

Our provided exercise illustrates probability without replacement beautifully—the chips are drawn from the urns and not replaced, altering the possible outcomes with each draw. This aspect is critical in figuring out the likelihood probabilities because it affects the number of possible combinations of chips that can be drawn from the urns, as clearly denoted in the solution using combinatorial formulas \( (\frac{{n \choose k}}{{m \choose k}}) \), where \( n \) and \( m \) represent the total and subset numbers, respectively.

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