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In a semiconductor manufacturing process, three wafers from a lot are tested. Each wafer is classified as pass or fail. Assume that the probability that a wafer passes the test is 0.8 and that wafers are independent. Determine the probability mass function of the number of wafers from a lot that pass the test.

Short Answer

Expert verified
The PMF is: \( P(X=0) = 0.008 \), \( P(X=1) = 0.096 \), \( P(X=2) = 0.384 \), \( P(X=3) = 0.512 \).

Step by step solution

01

Define the Random Variable

Let the random variable \( X \) represent the number of wafers that pass the test out of three. \( X \) can take on the values 0, 1, 2, or 3, since there are three wafers tested.
02

Identify the Probability Distribution

The situation described involves independent trials with two outcomes (pass or fail) and a fixed number of trials (three wafers), fitting the criteria for a binomial distribution. Hence, \( X \sim \text{Binomial}(n=3, p=0.8) \).
03

Calculate the Probability of 0 Passing Wafers

Use the binomial probability formula: \[ P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} \] for \( k=0 \). So, \( P(X=0) = \binom{3}{0} (0.8)^0 (0.2)^3 = 1 \times 1 \times 0.008 = 0.008 \).
04

Calculate the Probability of 1 Passing Wafer

For \( k=1 \), \( P(X=1) = \binom{3}{1} (0.8)^1 (0.2)^2 = 3 \times 0.8 \times 0.04 = 0.096 \).
05

Calculate the Probability of 2 Passing Wafers

For \( k=2 \), \( P(X=2) = \binom{3}{2} (0.8)^2 (0.2)^1 = 3 \times 0.64 \times 0.2 = 0.384 \).
06

Calculate the Probability of 3 Passing Wafers

For \( k=3 \), \( P(X=3) = \binom{3}{3} (0.8)^3 (0.2)^0 = 1 \times 0.512 \times 1 = 0.512 \).
07

Write the Probability Mass Function

The PMF of \( X \) is given by: \[ P(X=0) = 0.008, \quad P(X=1) = 0.096, \quad P(X=2) = 0.384, \quad P(X=3) = 0.512 \].

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Key Concepts

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

Probability Mass Function
The probability mass function (PMF) is a fundamental concept in probability and statistics. It is a function that provides the probabilities of different outcomes for a discrete random variable. In simple terms, it tells us how the probabilities are distributed over various possible values a random variable can take. For example, in our semiconductor manufacturing problem, the random variable is the number of wafers passing the test. The PMF indicates the likelihood of having 0, 1, 2, or 3 wafers passing out of the tested three. The values are calculated using the binomial formula:
  • For 0 passing wafers: \( P(X=0) = 0.008 \).
  • For 1 passing wafer: \( P(X=1) = 0.096 \).
  • For 2 passing wafers: \( P(X=2) = 0.384 \).
  • For 3 passing wafers: \( P(X=3) = 0.512 \).
These probabilities tell us how likely each scenario is, which can help in making informed decisions or predictions. Every possible outcome's probability totals to 1, confirming that all possible scenarios are accounted for.
Random Variable
A random variable is a core concept that allows us to organize and quantify outcomes in a probability distribution. It is essentially a variable whose values are numerical outcomes of a random phenomenon. In our context, the random variable \( X \) represents the number of wafers that pass the test from a given sample of three.

Random variables can be either discrete or continuous. Here, \( X \) is discrete because it takes on a countable number of values: 0, 1, 2, or 3. Each number reflects a possible outcome of the testing process. Understanding random variables is crucial because they enable us to use mathematical tools, like the binomial distribution, to analyze and predict outcomes effectively.

In this problem, we express the random variable's possible outcomes through the probability mass function, which showcases the chances of each discrete outcome. This way, random variables help bridge the gap between real-world randomness and mathematical theory.
Independent Trials
Independent trials refer to the repeated execution of an experiment, where the result of each trial does not affect the results of others. This concept is vital in understanding how probability works in repeated scenarios. For instance, the probability of each wafer passing the test is given as 0.8 in this exercise. Each wafer's result is independent of the others, meaning:
  • The outcome of the first wafer does not influence the outcome of the second or third wafers.
  • Each wafer has the same chance of passing, unaffected by the outcomes of the other wafers.
This assumption of independence is key to applying the binomial distribution model, as it ensures that the basic probability for a single wafer passing remains constant across trials.

Moreover, independent trials help simplify calculations because probabilities of individual events can be multiplied to find overall outcomes. If trials were not independent, calculating probabilities would require more complex statistical models to account for potential interdependencies.

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