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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 of the number of passing wafers is 0.008, 0.096, 0.384, and 0.512 for 0, 1, 2, and 3 passing wafers, respectively.

Step by step solution

01

Define the Random Variable

Let the random variable \(X\) represent the number of wafers that pass the test. Here, \(X\) can take on the values 0, 1, 2, or 3, since there are three wafers and each one can either pass or fail.
02

Identify the Probability Distribution

Since each wafer is independent and has two possible outcomes (pass or fail), \(X\) follows a binomial distribution. The number of trials \(n = 3\) and the probability of success (passing) on each trial \(p = 0.8\).
03

Write the Probability Mass Function Formula

The probability mass function (PMF) for a binomial distribution is given by:\[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]where \(n\) is the number of trials, \(k\) is the number of successes, and \(p\) is the probability of success on each trial.
04

Calculate the Probability for Each Outcome

We calculate \(P(X = k)\) for \(k = 0, 1, 2, 3\):\(P(X = 0) = \binom{3}{0} (0.8)^0 (0.2)^3 = 0.008\)\(P(X = 1) = \binom{3}{1} (0.8)^1 (0.2)^2 = 0.096\)\(P(X = 2) = \binom{3}{2} (0.8)^2 (0.2)^1 = 0.384\)\(P(X = 3) = \binom{3}{3} (0.8)^3 (0.2)^0 = 0.512\)
05

Compile the Probability Mass Function

The probability mass function is:\(P(X = 0) = 0.008\)\(P(X = 1) = 0.096\)\(P(X = 2) = 0.384\)\(P(X = 3) = 0.512\)

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Key Concepts

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

Probability Distribution
A probability distribution is like a roadmap that tells us how likely different outcomes of an experiment are. Imagine you're rolling a die. Each number has a chance of landing face up, which represents a simple probability distribution, where each of the six outcomes is equally likely. In the case of the semiconductor manufacturing process from our exercise, it is evident we are dealing with a binomial distribution. This is because we are interested in the probability of a certain number of successes (wafers passing) in a fixed number of trials (testing three wafers).

For any binomial probability distribution, two key parameters define it:
  • Number of trials: In our case, it's the three wafers being tested for pass or fail.
  • Probability of success in each trial: Here, it is the 0.8 chance for a wafer to pass the test.

By evaluating these components, we can understand how the probability of observing different numbers of successful outcomes is structured.
Probability Mass Function
The probability mass function, or PMF, is a tool that helps us understand the distribution of discrete random variables. It's like a guide that tells you the probability that a specific discrete outcome will occur. For example, if you flip a coin twice, the PMF can help you understand the likelihood of getting zero, one, or two heads.

In our semiconductor scenario, the probability mass function is used to find the probability that a certain number of wafers out of three pass the test. We define the PMF for a binomial distribution with the formula:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] - **Where:**
  • \(n\) is the total number of trials (in our case, 3 wafers).
  • \(k\) is the number of successes (number of wafers passing).
  • \(p\) denotes the probability of success on a single trial (here, 0.8).

The PMF tells us, for example, that there's a high probability (0.512) all three wafers pass, and a chance (0.008) none do, helping both engineers and statisticians predict outcomes and make informed decisions.
Random Variables
Random variables might sound intimidating, but they're just a way of linking numbers to real-world outcomes. For example, one could say that the number on the face of a die after a roll is a random variable. In the semiconductor exercise, the random variable \(X\) represents the number of wafers that pass the test.

The concept of random variables is crucial because it allows us to translate outcomes into mathematical terms, helping us to calculate probabilities. Random variables can be **discrete** or **continuous**. In our case, since we are counting the number of wafers passing out of three, \(X\) is a discrete random variable since it can only take on whole number values (0, 1, 2, or 3 in this example).

Understanding random variables is foundational to exploring probability distributions and probability mass functions, as it allows us to frame real-world situations in a mathematically precise way, providing deeper insights into the likelihood of different outcomes.

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Most popular questions from this chapter

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