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Question

12) One store notes that the probability of some type of error in a telephone order is 0.2. A supervisor randomly selects telephone orders and carefully inspects each one.

1. (2pts) What is the probability that the third telephone order selected will be the first to contain an error?
2. (3pts) What is the probability that the supervisor will inspect between two and six (inclusive) telephone orders before finding an error? 
3. (3pts) What is the probability that the inspector will examine at least seven orders before finding an error?
4. (4pts) Suppose the first four telephone orders contain no errors, what is the probability that the first error will be on the eighth order or later?

Answer 1

Each inspection could have 2 results: good (no error) or bad (error).

P(bad) = 0.2, P(good) = 1 - 0.2 = 0.8.

It is assumed that results of inspections are independent.

(1) P(first error will be for the 3rd order) = P(good, good, bad) = 0.8*0.8*0.2 = 0.128

It is easier to solve (3) the next:

(3) P(the inspector will examine at least seven orders) = P(the first 6 orders are good) = (0.8)⁶ = 0.262144

Now solve (2).

(2) Consider 2 random events:

A: the inspector will find error in the 1st order

B: the inspector will find the first error between the 2nd and 6th order (inclusive)

C: the inspector will find the first error after the 6th order.

These events are mutually exclusive and one of them must occur.

So, P(A) +P(B) + P(C) = 1

P(A) = 0.2

P(C) = P(the first 6 orders are good) = (0.8)⁶

P(B) = 1 -P(A) - P(C) = 1 - 0.2 - (0.8)⁶ = 0.537856

(4) P(the first error will be on the eighth order or later | the first four telephone orders contain no errors) =

= P(the 5th, 6th, and 7th orders are good | the first 4 orders are good)

Because we assume independence, results of following inspections do not depend on the previous results:

P(the 5th, 6th, and 7th orders are good | the first 4 orders are good) = P(the 5th, 6th, and 7th orders are good) = (0.8)³ = 0.512.