Question 1164737: A trading company has eight computers that it uses 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 in the same day
compute the solution in R ?
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(2189)   (Show Source):
You can put this solution on YOUR website! To solve this, we will use the properties of the **Binomial Distribution** (for successes in a fixed number of trials) and the **Geometric Distribution** (for the time until an event occurs).
### Mathematical Breakdown
**Part (a): Probability that all eight computers fail in a day**
Since each computer is an independent trial with a constant probability of failure , the number of failures follows a Binomial distribution: .
The probability of all 8 failing is:
**Part (b): Mean number of days until a specific computer fails**
The number of days until a specific event occurs follows a Geometric distribution. The mean (expected value) is given by:
**Part (c): Mean number of days until all eight computers fail on the same day**
In this case, we treat the "event" as the simultaneous failure of all 8 computers. Let the probability found in part (a) be . The mean number of days until this specific event occurs is:
---
### R Solution
You can copy and paste the following code into your R console or RStudio:
```r
# Define parameters
n <- 8
p <- 0.005
# (a) Probability that all eight computers fail in a day
# Using the dbinom function or direct calculation
prob_all_fail <- dbinom(8, size = n, prob = p)
# prob_all_fail <- p^8
cat("(a) Probability all 8 fail in a day:", format(prob_all_fail, scientific = TRUE), "\n")
# (b) Mean number of days until a specific computer fails
# Mean of Geometric distribution is 1/p
mean_days_specific <- 1 / p
cat("(b) Mean days until a specific computer fails:", mean_days_specific, "days\n")
# (c) Mean number of days until all eight computers fail in the same day
# Mean is 1 over the probability of the event occurring
mean_days_all_fail <- 1 / prob_all_fail
cat("(c) Mean days until all 8 fail on same day:", format(mean_days_all_fail, big.mark = ","), "days\n")
```
### Explanation of Results
* **(a)** The probability is . This is effectively zero, making it an extremely rare "black swan" event.
* **(b)** On average, you can expect a specific computer to fail once every **200 days**.
* **(c)** The mean wait time for all 8 to fail at once is ** days**. For perspective, the universe is approximately days old, so this event is statistically unlikely to ever happen in human history.
Would you like to see how the probability changes if the computers were not independent (e.g., a power surge affecting all of them at once)?
Answer by ikleyn(53618)  (Show Source):
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