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In October, 1994, a flaw in a certain Pentium chip installed in computers was discovered that could result in a wrong answer when performing a division. The manufacturer initially claimed that the chance of any particular division being incorrect was only 1 in 9 billion, so that it would take thousands of years before a typical user encountered a mistake. However, statisticians are not typical users; some modern statistical techniques are so computationally intensive that a billion divisions over a short time period is not outside the realm of possibility. Assuming that the 1 in 9 billion figure is correct and that results of different divisions are independent of one another, what is the probability that at least one error occurs in one billion divisions with this chip?

Step by step solution

01

Identify Parameters

Let's denote the probability of a single division being incorrect as \( p \). According to the problem, \( p = \frac{1}{9,000,000,000} \). We want to find the probability that there is at least one error in one billion divisions.

02

Apply Complement Rule

To find the probability of at least one error, we'll use the complement rule. First, calculate the probability of no errors. The probability of a correct division is \( 1-p = \frac{8,999,999,999}{9,000,000,000} \). For one billion independent divisions, the probability that all divisions are correct (i.e., no errors) is \( \left(1-p\right)^{1,000,000,000} \).

03

Calculate Probability of No Errors

Calculate \( \left(1-p\right)^{1,000,000,000} \) using the approximation for small \( p \), which is \( e^{-np} \), where \( n \) is the number of trials and \( p \) the probability of error in one trial. Thus, \( e^{-1,000,000,000 \times \frac{1}{9,000,000,000}} = e^{-\frac{1}{9}} \).

04

Find Probability of at Least One Error

Use the complement rule to determine the probability of at least one error: \( 1 - e^{-\frac{1}{9}} \). Calculating this gives approximately \( 1 - 0.8948 \approx 0.1052 \).

05

Conclusion

The probability that at least one error occurs in one billion divisions is approximately 0.1052, or 10.52%.

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Key Concepts

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

Complement Rule

In probability theory, the complement rule is a simple yet powerful tool to solve many problems with ease. This concept is all about working backwards.
Instead of directly calculating the probability of an event, we find the probability of it not happening and subtract it from 1.

For example, it might be tough to calculate the likelihood of getting at least one error in a billion computer operations. However, it is much easier to find the probability of no errors at all.
The complement rule tells us that the probability of at least one error is equal to one minus the probability of no errors.

• If the chance of an error in a single operation is tiny, as in our original problem, we use: \( \text{Probability of at least one error} = 1 - P(\text{no errors in } n \text{ trials}) \).

Understanding and using the complement can save you from tedious calculations and open a clearer path to your answer.

Independent Events

When events are considered independent, the result of one does not affect the outcome of the other.
In the context of our original problem, each computer division is treated as an independent event.

• This means that whether one division results in an error does not impact the likelihood of subsequent divisions erring.

To calculate the probability for such scenarios, you multiply the probabilities of individual events.

For instance, to find the probability that no errors occur in a sequence of divisions, you multiply the probability of a correct result (one minus the error probability) for each division:
\( P(\text{no errors in } n \text{ trials}) = (1-p)^n \).

Understanding the concept of independent events is crucial in the realm of probability, as it frequently simplifies problems by allowing calculations one step at a time.

Approximation Techniques

Approximation techniques are incredibly useful when dealing with complex or large calculations in probability.
In our problem, the error probability is so small, and the number of trials so large, that directly calculating \( (1-p)^n \) becomes unwieldy.

• Here, an approximation method known as the exponential approximation is used, specifically \( e^{-np} \).

This stems from a fundamental result in probability, where when \( p \) is extremely small and \( n \) is large, \( (1-p)^n \) can be approximated by an exponent:
\( e^{-np} = e^{\text(Some small value close to zero)} \).

This simplification is derived from calculus and exponential functions providing smoother calculations involving vast numbers and extremely tiny probabilities.

This technique not only makes complex computations manageable but also illustrates how theory meets practical application in statistics.