3 CPA has found that 9 of 10 company audits contain substant

3. CPA has found that 9 of 10 company audits contain substantial errors. If the CPA audits a series of company accounts, what is the probability that the first account containing substantial errors?

A. is a third one to be audited?

B. Will occur on or after the third audited account?

4. We want to select 6 people from a class of 50. There are three girls in the class.

A) Number of ways of selecting 6 from the class

B) What is the probability of getting exactly two girls in the group?

Solution

Here we have a binomial situation with p=0.9 since nine out of ten (=9/10=0.9) records contain substantial errors.

If you want the third record to be the first one to have an error, then the first two contain no error (10% chance) so:

0.1*0.1*0.9 = 0.009 = 0.9%

The most easiest way to find out if it will occur on the third or afterwards, you need to subtract the possibilities that it is the first or second one examined:

First: 0.9
Second: 0.1*0.9 = 0.09
First or second: 0.9 + 0.09 = 0.99
Later: 1-0.99 = 0.01 = 1%

4 solution: We want to select 6 people from a class of 50. There are three girls in the class.

3 girls + 47 boys = total 50

A) Number of ways of selecting 6 from the class.

Lets use here combination concept.

A formula for the number of possible combinations of r

objects from a set of n objects. This is written in any of the ways shown below.

C(n ,r) Therefore, We have Number of ways selecting 6 from 50 = C(50 , 6 ) = 15 ,890, 700

P[of getting exactly two girls in the group]

i.e. selct 2 girls from total 3 girls and remaining 4 boys from 47

P(E) = n(E)/n(S)

where E is event getting exactly two girls, n(E) = total number of favourable outcomes for event E, n(S) = all possible distinct outcomes in sample space.

P(E) = C(3 ,2) + C(47 , 4 ) 3 + 178,365

----------------------- = ----------------------- = 0.01122

C( 50 , 6 ) 15 ,890 ,700