Question

The probability of a successful rocket launching equals 0.8. Suppose that launching attempts are made until 3 successful launchings have occurred. (a) What is the probability that exactly 6 attempts will be necessary? (b) What is the probability that fewer than 6 attempts will be required? (c) Suppose that each launching attempt costs $5000. In addition, a launching failure results in an additional cost of $500. Evaluate the expected cost.

ans: (a), 0.0409; (b), 0.983; (c), $19,125

Answer 1

here this follows negative binomial distribution with parameter r =3 and p=0.8

a)

P( exactly 6 attempts will be necessary )=P(exactly 2 success in first 5 and 3rd success in 6th)

=(5C2)*(0.8)^3*(0.2)^3 =0.04096~ 0.0410

b)P(X<6) =P(X=5)+P(X=4)+P(X=3)

=(4C2)*(0.8)^3*(0.2)^2+(3C2)*(0.8)^3*(0.2)^1+(0.8)^3

=0.9421 (if it icludes 6 th attempt, then probability will be 0.9421+0.0410 =0.9830)

c)

expected number of launches required =r/p=3/0.8=3.75

therefore expected cost =cost of 3 success +cost of 0.75 faliure

=3*(5000)+0.75*(5000+500)

=19125