A sample of 5 accounts is chosen without replacement from a set of 20. From past experience it is known that 20% of the accounts are in error. What is the probability that not more than two errors will occur in the sample?
Hello, matty888!
A sample of 5 accounts is chosen without replacement from a set of 20.
From past experience it is known that 20% of the accounts are in error.
What is the probability that not more than two errors will occur in the sample?
There are 20 accounts: .$\displaystyle \begin{array}{cc}\text{16 are good (G)} \\ \text{4 have errors (E)} \end{array}$
There are: .$\displaystyle {20\choose5} \:=\:15,504$ possible samples.
We want the probability that there are:
. . 0 errors or 1 error or 2 errors.
0 E and 5 G: .$\displaystyle {4\choose0}{16\choose5} \:=\:4368$ ways.
1 E and 4 G: .$\displaystyle {4\choose1}{16\choose4} \:=\:1820$ ways.
2 E and 3 G: .$\displaystyle {4\choose2}{16\choose3} \:=\:3360$ ways.
Hence, there are: .$\displaystyle 4368 + 1820 + 3360 \:=\:9548$ ways to get 0, 1 or 2 Errors.
Therefore: .$\displaystyle P(\text{0, 1 or 2 E}) \;=\;\frac{9548}{15,504} \;=\;\boxed{\frac{2387}{3876}}$