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An electronics assembly firm buys its microchips from three different suppliers; half of them are bought from firm \(X\), whilst firms \(Y\) and \(Z\) supply \(30 \%\) and \(20 \%\), respectively. The suppliers use different quality- control procedures and the percentages of defective chips are \(2 \%, 4 \%\) and \(4 \%\) for \(X, Y\) and \(Z\), respectively. The probabilities that a defective chip will fail two or more assembly-line tests are \(40 \%, 60 \%\) and \(80 \%\), respectively, whilst all defective chips have a \(10 \%\) chance of escaping detection. An assembler finds a chip that fails only one test. What is the probability that it came from supplier \(X\) ?

Short Answer

Expert verified
The probability that the chip came from supplier \(X\) given it only fails one test can be found by substituting \(P(Fails one test)\) in the formula obtained in step 3.

Step by step solution

01

Identify and Define the Known Probabilities

Firstly, a probability distribution for where the chips are coming from is given: \(P(X) = 0.5, P(Y) = 0.3, P(Z) = 0.2\). Here, \(X,Y,Z\) refer to different suppliers. Also, it is known that the probabilities of microchips being defective are \(P(Defect|X) = 0.02, P(Defect|Y) = 0.04, P(Defect|Z) = 0.04\). Lastly, the probabilities that a defective chip fails two or more tests are \(P(Fails 2+ tests|X) = 0.4, P(Fails 2+ tests|Y) = 0.6, P(Fails 2+ tests|Z) = 0.8\).
02

Calculate the Conditional Probabilities

It is asked to find the probability that a chip came from supplier \(X\) given it only fails one test. We can apply Bayes' Theorem here. However, we first need to calculate the probability of failures for one test regardless of the source. The probability of failing one test can be computed in two steps: \nFirst, by computing the probability of being defective regardless of the source and then by computing the probability of failing only one test given it is defective which is \(1 - P(Failing 2+ tests | Defect) = 1 - 0.1 = 0.9\). \nHence, \(P(Fails one test) = P(Defective) * P(Fails one test | Defective) = (P(X)*P(Defect|X) + P(Y)*P(Defect|Y) + P(Z)*P(Defect|Z)) * 0.9 = (0.5*0.02 + 0.3*0.04 + 0.2*0.04)*0.9\).
03

Apply Bayes' Theorem

To calculate the probability that a chip came from supplier \(X\) given that it only fails one test, Bayes' Theorem should be applied: \(P(X | Fails one test) = [P(X)*P(Defect|X)*P(Fails one test|Defective)] / P(Fails one test) = [0.5*0.02*0.9] / P(Fails one test)\).

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Key Concepts

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

Understanding Conditional Probability
Conditional probability is the likelihood of an event or outcome occurring, based on the occurrence of a previous event. In the context of our problem, we want to determine the probability that a microchip came from supplier \(X\) given that it only fails one test. This probability is represented as \(P(X | \text{Fails one test})\).

To calculate this, we utilize Bayes' Theorem, which provides a way to reverse conditional probabilities. It is expressed as:
  • \(P(A | B) = \frac{P(B | A) \times P(A)}{P(B)}\)
Here, \(A\) is the event we are interested in (chip from \(X\)), while \(B\) is the observed event (fails one test).

Bayes’ Theorem helps us update our probability assessment since it assumes prior beliefs (the known defectiveness rates and supplier probabilities) and updates them with new evidence (the chip failing one test). This theorem is a powerful tool in statistics, especially when dealing with probabilities that change based on new information.

In this scenario, conditional probability involves several known probabilities such as the likelihood of purchasing from each supplier and the defect rates from these suppliers. Understanding these can help in reasoning about which supplier a defective chip most likely originated from.
Exploring Probability Distribution
Probability distribution describes how the probabilities are distributed over various outcomes. In the given task, it outlines how the microchips are bought from different firms and helps us understand the likelihood of each firm being the source of a defective chip.

The probability distribution is given as:
  • \(P(X) = 0.5\) for supplier \(X\)
  • \(P(Y) = 0.3\) for supplier \(Y\)
  • \(P(Z) = 0.2\) for supplier \(Z\)
This tells us that half of the chips come from \(X\), while \(30\%\) and \(20\%\) come from \(Y\) and \(Z\) respectively.

This information is crucial when determining the overall probability of any chip being defective, as it forms the basis of the aggregate defect rate calculations across all suppliers. We essentially combine these supplier distribution probabilities with the defect rates to find out how probable it is for a defective chip to be sourced from a specific supplier.
Analyzing Defective Microchips
Defective microchips are a crucial aspect of the exercise. Each supplier has a specific rate of defective chips:
  • Supplier \(X\): \(2\%\) defective
  • Supplier \(Y\): \(4\%\) defective
  • Supplier \(Z\): \(4\%\) defective
These percentages tell us the quality control measure of each firm. However, it becomes even more interesting when considering the probability that defective chips escape detection. All defective chips have a \(10\%\) chance of being missed, affecting the probability of detecting defects.

For each supplier, the defective microchips that fail 2 or more tests can be categorized:
  • Fails 2+ tests: \(40\%\) for \(X\), \(60\%\) for \(Y\), \(80\%\) for \(Z\)
This affects how likely a chip is to fail only one test. Understanding these defective rates, along with other probabilities, helps in applying the distributions correctly in Bayesian calculations. These calculations allow us to pinpoint the probability of a specific supplier being the source of a failing microchip, given it only fails one test.

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Most popular questions from this chapter

A discrete random variable \(X\) takes integer values \(n=0,1, \ldots, N\) with probabilities \(p_{n}\). A second random variable \(Y\) is defined as \(Y=(X-\mu)^{2}\), where \(\mu\) is the expectation value of \(X\). Prove that the covariance of \(X\) and \(Y\) is given by $$ \operatorname{Cov}[X, Y]=\sum_{n=0}^{N} n^{3} p_{n}-3 \mu \sum_{n=0}^{N} n^{2} p_{n}+2 \mu^{3} $$ Now suppose that \(X\) takes all of its possible values with equal probability, and hence demonstrate that two random variables can be uncorrelated, even though one is defined in terms of the other.

A boy is selected at random from amongst the children belonging to families with \(n\) children. It is known that he has at least two sisters. Show that the probability that he has \(k-1\) brothers is $$ \frac{(n-1) !}{\left(2^{n-1}-n\right)(k-1) !(n-k) !} $$ for \(1 \leq k \leq n-2\) and zero for other values of \(k\). Assume that boys and girls are equally likely.

As every student of probability theory will know, Bayesylvania is awash with natives, not all of whom can be trusted to tell the truth, and lost, and apparently somewhat deaf, travellers who ask the same question several times in an attempt to get directions to the nearest village. One such traveller finds himself at a T-junction in an area populated by the Asciis and Bisciis in the ratio 11 to 5 . As is well known, the Biscii always lie, but the Ascii tell the truth three-quarters of the time, giving independent answers to all questions, even to immediately repeated ones. (a) The traveller asks one particular native twice whether he should go to the left or to the right to reach the local village. Each time he is told 'left'. Should he take this advice, and, if he does, what are his chances of reaching the village? (b) The traveller then asks the same native the same question a third time, and for a third time receives the answer 'left'. What should the traveller do now? Have his chances of finding the village been altered by asking the third question?

By shading or numbering Venn diagrams, determine which of the following are valid relationships between events. For those that are, prove the relationship using de Morgan's laws. (a) \(\overline{(\bar{X} \cup Y)}=X \cap \bar{Y}\). (b) \(\bar{X} \cup \bar{Y}=\overline{(X \cup Y)}\). (c) \((X \cup Y) \cap Z=(X \cup Z) \cap Y\). (d) \(X \cup \overline{(Y \cap Z)}=(X \cup \bar{Y}) \cap \bar{Z}\). (e) \(X \cup \overline{(Y \cap Z)}=(X \cup \bar{Y}) \cup \bar{Z}\).

As assistant to a celebrated and imperious newspaper proprietor, you are given the job of running a lottery, in which each of his five million readers will have an equal independent chance, \(p\), of winning a million pounds; you have the job of choosing \(p\). However, if nobody wins it will be bad for publicity, whilst if more than two readers do so, the prize cost will more than offset the profit from extra circulation - in either case you will be sacked! Show that, however you choose \(p\), there is more than a \(40 \%\) chance you will soon be clearing your desk.

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