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US A trading company uses eight computers to trade on the New York Stock Exchange (NYSE). The probability of a computer failing in a day is \(0.005,\) and the computers fail independently. Computers are repaired in the evening, and each day is an independent trial. a. What is the probability that all eight computers fail in a day? b. What is the mean number of days until a specific computer fails? c. What is the mean number of days until all eight computers fail on the same day?

Short Answer

Expert verified
a. Probability is \(3.90625 \times 10^{-23}\). b. Mean is 200 days. c. Mean is \(2.56 \times 10^{22}\) days.

Step by step solution

01

Understanding the Probability of All Computers Failing

To find the probability that all eight computers fail on a given day, we need to multiply the individual probabilities of each computer failing on that day. The probability for one computer to fail is \(0.005\). Hence, the probability that all eight computers fail is \((0.005)^8\).
02

Calculate the Probability in Step 1

Now, calculate the probability that all eight computers fail simultaneously on a single day. Use the formula from the previous step:\[(0.005)^8 = 3.90625 \times 10^{-23}\]
03

Understand the Problem in Part B

For part b, we need to find the mean number of days until a specific computer fails. This situation follows a geometric distribution where the probability of success (failure in this context) is \(0.005\).
04

Calculate the Mean Days Until One Computer Fails

The mean (expected value) for a geometric distribution is given by \(\frac{1}{p}\), where \(p\) is the probability of a single trial's success. In this case:\[\frac{1}{0.005} = 200\]Thus, the mean number of days until a specific computer fails is 200.
05

Understanding the Problem in Part C

For this part, we need to find the mean number of days until all eight computers fail on the same day. This follows the process for a geometric distribution again, where the probability from Step 1 is \(3.90625 \times 10^{-23}\).
06

Calculate the Mean Days Until All Computers Fail on the Same Day

Using the formula for the mean of a geometric distribution, \(\frac{1}{p}\), where \(p\) is the probability of all eight computers failing simultaneously from Step 2:\[\frac{1}{3.90625 \times 10^{-23}} \approx 2.56 \times 10^{22}\]This means it would take approximately \(2.56 \times 10^{22}\) days on average for all eight computers to fail simultaneously on the same day.

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Key Concepts

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

Probability Calculation
To calculate probabilities in scenarios where events occur independently, you can multiply the individual probabilities of each event. Let's consider the exercise where a trading company uses eight computers. The probability of a single computer failing in one day is given as \(0.005\).

The goal in the exercise was to determine the probability that all eight computers fail on the same day. Each computer acts independently, which means the probability of all of them failing can be found by raising the individual failure probability to the power of the number of computers. Therefore, the formula looks like this:
  • The probability for all eight computers failing: \[(0.005)^8 = 3.90625 \times 10^{-23}\]
This result allows us to understand the incredibly low likelihood that all computers will experience failure on the same day. This calculation is a direct application of the rule for independent probabilities.
Expected Value
Expected value, in simple terms, is the average expected outcome of a random event if it was repeated many times. In a geometric distribution, expected value helps us determine the average number of trials until an event—such as a computer failing—happens.

For a geometric distribution, which applies when events are independent and have equal probability each time, the expected number of trials (or days, in this context) until the first success (or failure) is calculated as:
  • \(\frac{1}{p}\)
Here, \(p\) is the probability of failure for one computer in a day, which is \(0.005\). Therefore, the mean number of days until a specific computer fails is \(\frac{1}{0.005} = 200\) days.

In another scenario from the exercise where we need all eight computers to fail simultaneously, the probability changes to \(3.90625 \times 10^{-23}\) (as calculated earlier). Thus, the expected mean number of days until all computers fail on the same day is:
  • \(\frac{1}{3.90625 \times 10^{-23}} \approx 2.56 \times 10^{22}\) days
This illustrates how rare simultaneous failures are, given the extremely high expected days calculated.
Independent Events
Understanding independent events is critical to accurately computing probabilities like we did in this exercise. An event is independent if the occurrence of any one event doesn't affect the probability of another event occurring. This property holds true for the computers in the trading company as each computer's operation is unaffected by the state of the others.

When dealing with independence:
  • If Event A and Event B are independent, the probability of both occurring is the product of their probabilities: \(P(A \cap B) = P(A) \times P(B)\).
For the trading company's situation, because each computer failing is an independent event, we used this rule to calculate the probability of simultaneous failures, multipliyed the failure probabilities for all eight computers.

Recognizing and applying the principle of independent events is crucial in probability calculations and allows us to analyze complex scenarios effectively.

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Most popular questions from this chapter

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