Probability

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library(knitr)
library(matlib)

Problem Set 1

The answers are in the following pages.

Problem Set 2

  1. Your organization owns a copier (future lawyers, etc.) or MRI (future doctors). This machine has a manufacturer’s expected lifetime of 10 years. This means that we expect one failure every ten years. (Include the probability statements and R Code for each part.).

  1. What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a geometric. (Hint: the probability is equivalent to not failing during the first 8 years..)

  2. What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as an exponential.

  3. What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a binomial. (Hint: 0 success in 8 years)

  4. What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a Poisson.

Answer

  1. Here, each trial is outcome in each year, which can be a success or a failure. Per manufacturer’s warranty, we expect 1 failure every 10 years. It’s a case of Geometric Distribution. Therefore, “probability of machine failing after 10 years” is equivalent to “1 - probability of machine not failing after 10 years”.

Let’s designate p_failure as “probability of failure 10 years” i.e. 1 failure in 10 years.
Let’s designate p_success = 1 - p_failure.
Let t be number of trials.
Let p_t_success be the probability of success in t years. It’s requied to find the probability of failure after 8 years. Therefore, the machine will succeed for 7 years.

p_failure <- 0.1
p_success <- 1 - p_failure
t <- 8
p_t_success <- ((p_success)^(t - 1)) * (p_failure)^(1)
paste("P(failure after 8 years) =", p_t_success)
## [1] "P(failure after 8 years) = 0.04782969"

Let E be the Expected Value, which is calculated below, using formula:

E <- 1/p_failure
paste("Expected Value =", E)
## [1] "Expected Value = 10"

Let sd be the Standard Deviation, which is calculated below, using formula:

sd <- sqrt((1 - p_failure) / (p_failure)^2)
paste("Standard Deviation =", sd)
## [1] "Standard Deviation = 9.48683298050514"
  1. The formula for Exponential Distribution is: f(x;Lambda) = (e^( -(Lambda * x) ).
    Let p_t_success_exp be the probability of success in t years, modeled in exponential manner.
    Let Lambda be the parameter for Exponential Distribution.
Lambda <- p_failure
p_t_success_exp <- ( exp(-(Lambda * t)) )
paste("P(t >= 8 years) =", p_t_success_exp)
## [1] "P(t >= 8 years) = 0.449328964117222"

Let E_exp be the Expected Value, modeled in exponential manner, which is calculated below, using formula:

E_exp <- 1/Lambda
paste("Expected Value =", E_exp)
## [1] "Expected Value = 10"

Let sd_exp be the Standard Deviation, modeled in exponential manner, which is calculated below, using formula:

sd_exp <- sqrt((1/Lambda^2))
paste("Standard Deviation =", sd_exp)
## [1] "Standard Deviation = 10"
  1. The formula for Binomial Distribution is: P(t > 8) = Choose(8, 0) * p_success^0 * p_failure^8.
    Let p_t_success_binom be the probability of success in t years, modeled in binomial manner.
p_t_success_binom <- choose(8, 0) * ((p_failure)^0) * ((p_success)^8)
paste("P(t >= 8 years) =", p_t_success_binom)
## [1] "P(t >= 8 years) = 0.43046721"

Let E_binom be the Expected Value, modeled in binommial manner, which is calculated below, using formula:

E_binom <- t * p_failure
paste("Expected Value =", E_binom)
## [1] "Expected Value = 0.8"

Let sd_binom be the Standard Deviation, modeled in binommial manner, which is calculated below, using formula:

sd_binom <- sqrt(t * p_failure * p_success)
paste("Standard Deviation =", sd_binom)
## [1] "Standard Deviation = 0.848528137423857"
  1. The formula for Poisson Distribution is: P(t = 8) = (Lambda^8) * e^(-Lambda)/8!, where Lambda average of outcomes i.e. Expected Value, which we computed earlier as 10.
    Let p_t_success_poisson be the probability of success in t years, modeled in Poisson manner.
Lambda <- 10
p_t_success_poisson <- (exp(-1 * Lambda) * Lambda^8) / factorial(8)
paste("P(t >= 8 years) =", p_t_success_poisson)
## [1] "P(t >= 8 years) = 0.11259903214902"

Let E_poisson, which we used before:

E_poisson <- Lambda
paste("Expected Value =", E_poisson)
## [1] "Expected Value = 10"

Let sd_poisson be the Standard Deviation, modeled in Poisson manner, which is calculated below, using formula:

sd_Poisson <- sqrt(Lambda)
paste("Standard Deviation =", sd_Poisson)
## [1] "Standard Deviation = 3.16227766016838"

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