Question 1164741: 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 ?
Answer by CPhill(2189)   (Show Source):
You can put this solution on YOUR website! This problem involves two different probability distributions: the **Binomial Distribution** (for the number of computers failing in a single day) and the **Geometric Distribution** (for the number of days until a specific event occurs).
### (a) Probability that all eight computers fail in a day
Since the computers fail independently and each has the same probability of failure, we use the Binomial probability formula .
For "all eight fail," we have and :
### (b) Mean number of days until a specific computer fails
The number of trials (days) until the first "success" (failure of the computer) follows a Geometric Distribution. The mean (expected value) of a geometric distribution is .
### (c) Mean number of days until all eight computers fail on the same day
This is another Geometric Distribution problem, but the "success" event is now the extremely rare occurrence of all eight computers failing simultaneously.
1. First, find the probability of the event calculated in part (a).
2. The mean days until this occurs is .
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### R Code Solution
You can run the following code in R to get the exact values:
```r
# Parameters
n_computers <- 8
p_fail <- 0.005
# (a) Probability all 8 fail in a single day
# We use dbinom(k, size, prob)
prob_all_fail_day <- dbinom(8, size = n_computers, prob = p_fail)
# Or simply p_fail^8
# prob_all_fail_day <- p_fail^8
print(paste("Prob all 8 fail in a day:", format(prob_all_fail_day, scientific = TRUE)))
# (b) Mean days until a specific computer fails
# E[X] = 1/p
mean_days_specific <- 1 / p_fail
print(paste("Mean days until a specific computer fails:", mean_days_specific))
# (c) Mean days until all 8 fail in the same day
# E[X] = 1 / P(all 8 fail)
mean_days_all_fail <- 1 / prob_all_fail_day
print(paste("Mean days until all 8 fail in the same day:", format(mean_days_all_fail, big.mark = ",")))
```
### Interpretation of Results
* **Part (a):** The probability is . This is infinitesimally small, meaning it is almost impossible for this to happen on any given day.
* **Part (b):** You can expect a specific computer to fail roughly once every **200 days**.
* **Part (c):** You can expect all eight computers to fail on the same day once every ** days**. For context, this is significantly longer than the age of the universe.
Would you like me to calculate the probability that **at least one** computer fails in a day?
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