SOLUTION: Task 21. Two auditors check the accounts of the company. Each randomly selected 8 accounts. It is known that the probability of an erroneous account among company accounts is 0.05.

Question 1160049: Task 21. Two auditors check the accounts of the company. Each randomly selected 8 accounts. It is known that the probability of an erroneous account among company accounts is 0.05. What is the likelihood that: a. each auditor finds at least one erroneous account; b. will at least one erroneous account be detected?
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
p(a) finding at least 1 is 1-p(finding 0)=1-0.95^8=0.3366
That is the same for b.
They both have to find at least 1, and that is the joint probability of 0.3366^2=0.1133 rounding at end.
Sixteen accounts are checked randomly. The probability of finding at least 1 is 1-p(finding 0)=1-0.95^16=0.5599
This makes qualitative sense in that the probability of both finding at least 1 is more restrictive than just one auditor's finding at least 1.
RELATED QUESTIONS
I saw the below questions but cannot see the answers A company reports 12% of its... (answered by ikleyn)

In a local cellular phone area, company A accounts for 70% of the cellular phone market,... (answered by scott8148)

I am having a problem of putting the following into an linear equation. A company... (answered by htmentor)

Of 41 bank accounts at small bank, 21 accounts have values of less than $1000 and the... (answered by Fombitz)

The cellular spinoff company Jog wants to estimate the proportion of households that... (answered by stanbon)

A study of sample of 196 bank accounts showed the average size of bank accounts is Rs.... (answered by Boreal)

in a cellular phone area, company A accounts for 70% of the cellular phone market, while... (answered by robertb)

assume that a certain company was producing a bottles of soft drinks it is known that the (answered by Boreal)

Of 41 bank accounts at small bank, 21 accounts have values of less than $1000 and the... (answered by Fombitz)