Question

A IRS auditor randomly selects 3 tax returns from 60 returns of which 15 contain errors. what is the probability that she selects none of those containing errors?

A) 0.4219 B)0.4147 C) 0.0156 D)0.0133

Answer 1

From the 60 tax returns 15 contain errors.

That is (60-15) = 45 do not contain errors.

P(none contain errors) = Probability that all 3 tax returns selected from 45 tax returns which do

not contain errors.

So,

3 tax returns selected from 45 tax returns by ⁴⁵C₃ ways.

Total number of ways to select 3 tax returns from 60 = ⁶⁰C₃

Therefore,

P( None of 3 containing errors) = ⁴⁵C₃ / ⁶⁰C₃

In general,

ⁿC_r = n! / [ (n-r)! *r! ]

so ,

⁴⁵C₃ = 45! / [ (45 - 3)! * 3!]

= 45! / (42! * 3!)

= 14190

And

⁶⁰C₃ = 60! / [ (60-3)! * 3! ]

= 60! / 57! * 3!

= 34220

Therefore,

P( None of 3 containing errors) = ⁴⁵C₃ / ⁶⁰C₃

= 14190 / 34220

= 0.4147