Three cooks X, Y and Z bake a special kind of cake, and with respective probabilities 0.02, 0.03 and 0.05, it fails to rise. In the restaurant where they work, X bakes 50%, Y bakes 30% and Z bakes 20% of cakes. What is the proportion of failures caused by X?
The question asks us to find the proportion of failed cakes that were baked by cook X. We are given the individual probabilities of failure for each cook (X, Y, and Z) and the proportion of total cakes each cook bakes. This type of problem requires us to use concepts from probability, specifically the Law of Total Probability and Bayes' Theorem.
Let's list the probabilities provided in the question:
Notice that the proportions of cakes baked by each cook sum up to \(0.50 + 0.30 + 0.20 = 1.00\), covering all cakes baked in the restaurant.
Before we can find the proportion of failures caused by X, we need to determine the overall probability of a cake failing, regardless of who baked it. This is calculated using the Law of Total Probability. The total probability of an event (failure, F) is the sum of the probabilities of that event occurring under each possible condition (being baked by X, Y, or Z), weighted by the probability of each condition.
The formula for the Law of Total Probability is:
\(P(F) = P(F|X)P(X) + P(F|Y)P(Y) + P(F|Z)P(Z)\)
Let's plug in the given values:
\(P(F) = (0.02 \times 0.50) + (0.03 \times 0.30) + (0.05 \times 0.20)\)
Performing the multiplications:
\(P(F) = 0.010 + 0.009 + 0.010\)
Summing these values:
\(P(F) = 0.029\)
So, the total probability of a cake failing is 0.029, or 2.9%.
The question asks for the "proportion of failures caused by X". In probability terms, this is the conditional probability that a cake was baked by X GIVEN that it failed. This is denoted as \(P(X|F)\). We can calculate this using Bayes' Theorem.
Bayes' Theorem states:
\(P(X|F) = \dfrac{P(F|X) \times P(X)}{P(F)}\)
We have already calculated all the components needed for this formula:
Now, let's substitute these values into Bayes' Theorem:
\(P(X|F) = \dfrac{0.02 \times 0.50}{0.029}\)
Calculate the numerator:
\(0.02 \times 0.50 = 0.010\)
So, the equation becomes:
\(P(X|F) = \dfrac{0.010}{0.029}\)
To express this as a fraction, we can multiply the numerator and denominator by 1000 to remove the decimals:
\(P(X|F) = \dfrac{0.010 \times 1000}{0.029 \times 1000} = \dfrac{10}{29}\)
Therefore, the proportion of failures caused by X is \(\dfrac{10}{29}\).
| Cook | Proportion Baked (\(P(\text{Cook})\)) | Failure Probability (\(P(F|\text{Cook})\)) | Contribution to Total Failure (\(P(F|\text{Cook}) \times P(\text{Cook})\)) |
|---|---|---|---|
| X | 0.50 | 0.02 | \(0.50 \times 0.02 = 0.010\) |
| Y | 0.30 | 0.03 | \(0.30 \times 0.03 = 0.009\) |
| Z | 0.20 | 0.05 | \(0.20 \times 0.05 = 0.010\) |
Total probability of failure, \(P(F) = 0.010 + 0.009 + 0.010 = 0.029\).
Proportion of failures caused by X, \(P(X|F) = \dfrac{\text{Contribution of X to Failure}}{\text{Total Probability of Failure}} = \dfrac{0.010}{0.029} = \dfrac{10}{29}\).
Based on the probabilities provided and applying Bayes' Theorem, the proportion of cake failures that were baked by cook X is \(\dfrac{10}{29}\).
| Concept | Description | Formula Example (from this problem) |
|---|---|---|
| Conditional Probability | The probability of an event occurring given that another event has already occurred. | \(P(F|X)\): Probability of failure given it was baked by X. |
| Law of Total Probability | Used to find the total probability of an event when there are multiple mutually exclusive ways for the event to occur. | \(P(F) = \sum P(F|\text{Cook})P(\text{Cook})\) |
| Bayes' Theorem | Used to find the conditional probability of an event (e.g., Cook X) given the occurrence of another event (e.g., Failure), often used to update probabilities based on new evidence. | \(P(X|F) = \dfrac{P(F|X)P(X)}{P(F)}\) |
Bayes' Theorem is a fundamental concept in probability with wide-ranging applications. It's used in many fields to update beliefs or probabilities based on new data or evidence. Here are a few examples:
In essence, Bayes' Theorem helps us reverse conditional probabilities – going from the probability of an outcome given a cause, to the probability of a cause given an outcome.
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