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### HW1 Q4
### T ~ Exponential(lambda) = days from admission to discharge.
### Given: P(T <= 4) = 0.6. Find: P(T < 10 | T > 6).
### ! Rscript "/mnt/c/Users/kirch/OneDrive/Documents/dev/ut-11-hdi/docs/part1/q4.R"
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# Solve for lambda from F(4) = 1 - exp(-4*lambda) = 0.6
lambda = -log(1 - 0.6) / 4
cat("lambda-hat solved from P(T<=4)=0.6 :", lambda, "\n")

S = function(t) exp(-lambda * t)

cat("P(T<10 | T>6) direct  = (S(6)-S(10))/S(6) :", (S(6) - S(10)) / S(6), "\n")
cat("P(T<10 | T>6) memoryless = P(T<=4)        :", 1 - S(4), "\n")
cat("pexp(4, rate=lambda) check                :", pexp(4, rate = lambda), "\n")
cat("ANSWER (2 dp)                             :", round((S(6) - S(10)) / S(6), 2), "\n")