If Y is the minimum of the Xi’s and each Xi has k possibilities, then the total # of possibilities for X1, X2,…Xn is \(k^n\) , which becomes our denominator. The number of options we now know, so we need to find the number where no Xi’s = 1, which would become \((k-1)^n\)To get Y=1, \[\frac{(k-1+1)^n-(k-1)^n}{k^n}\] or to generalize whenever Y=x, \[\frac{(k-x+1)^n-(k-x)^n}{k^n}\]
#2. 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
#a. 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..)
p_fail <- 0.1 #once in 10 years
num_years <- 8
#probability
p_geom <- 1-pgeom(num_years-1,p_fail)
p_geom## [1] 0.4304672#expected value
geom_expected <- 1/p_fail
geom_expected## [1] 10#SD
geom_sd = sqrt((1-p_fail)/(p_fail)**2)
geom_sd## [1] 9.486833#b. What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as an exponential.
#prob
p_exp <- pexp(num_years,p_fail,lower.tail=FALSE)
p_exp## [1] 0.449329#EV
exp_expected <- 1/p_fail
exp_expected## [1] 10#SD
exp_SD <- sqrt(1/p_fail**2)
exp_SD## [1] 10#c. 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)
q <- 1-p_fail
k <-0
#prob
p_bin <- dbinom(k,num_years,p_fail)
p_bin## [1] 0.4304672#EV
bin_expected <- num_years*p_fail
bin_expected## [1] 0.8#SD
bin_SD <- sqrt(num_years*p_fail*q)
bin_SD## [1] 0.8485281# 8. What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a Poisson.
#prob
p_pois <- ppois(k,8/10) #avg number of failures in 8 years will be 8/10 of 1 or 0.8
p_pois## [1] 0.449329#EV
pois_expected <- 8/10
pois_expected## [1] 0.8#SD
pois_SD <- sqrt(8/10)
pois_SD## [1] 0.8944272