One out of every 119 tax returns that a tax auditor examines requires an audit. Find the probability that (a) the first return requiring an audit is the 25th return the tax auditor examines, (b) the first return requiring an audit is the first or second return the tax auditor examines, and (c) none of the first five returns the tax auditor examines require an audit.

A) 0.00686

B) 0.01673

C) 0.99587

Step-by-step explanation:

For this question, we are interested in the first success and we will Thus make use of geometric probability distribution.

It goes by the formula;

P(x) = [(1 - p)^(x - 1)]p

We are told that One out of every 119 tax returns that a tax auditor examines requires an audit.

Thus; p = 1/119 = 0.0084

A) Probability that the first return requiring an audit is the 25th return the tax auditor examines is given by;

P(25) = [(1 - 0.0084)^(25 - 1)]0.0084

P(25) = 0.00686

B) Probability the first requiring an audit is the first or second return the tax auditor examines, is;

P(1 or 2) = P(1) + P(2)

P(1) = [(1 - 0.0084)^(1 - 1)]0.0084

P(1) = 0.0084

P(2) = [(1 - 0.0084)^(2 - 1)]0.0084

P(2) = 0.00833

Thus;

P(1 or 2) = 0.0084 + 0.00833

P(1 or 2) = 0.01673

C) Probability that none of the first five returns the tax auditor examines require an audit is given by;

P(X > 5) = 1 - P(X ≤ 5)

Now,

P(X ≤ 5) is expressed as;

P(X ≤ 5) = P(1) + P(2) + P(3) + P(4) + P(5)

We know P(1) & P(2) already. Let's find P(3), P(4) and P(5)

P(3) = [(1 - 0.0084)^(3 - 1)]0.0084

P(3) = 0.00826

P(4) = [(1 - 0.0084)^(4 - 1)]0.0084

P(4) = 0.00819

P(5) = [(1 - 0.0084)^(5 - 1)]0.0084

P(5) = 0.00812

P(X ≤ 5) = 0.0084 + 0.00833 + 0.00826 + 0.00819 + 0.00812

P(X ≤ 5) = 0.00413

P(X > 5) = 1 - 0.00413

P(X > 5) = 0.99587

step-by-step explanation:

mean or average is 40.6

the answer is 99.