A box of 50 fuses contains five which are defective. If I take a sample of 20 out of the box, calculate the probability of there being at least two defective fuses.
Thanks
Hello, BabyMilo!
A box of 50 fuses contains five which are defective.
If I take a sample of 20 out of the box, calculate the probability
of there being at least two defective fuses.
There are: .$\displaystyle _{50}C_5 \:=\:{50\choose5}$ possible sampes.
We have: .$\displaystyle \begin{array}{c}\text{5 bad} \\ \text{45 good} \end{array}$
The opposite of "at least 2 bad" is "0 bad or 1 bad".
0 bad, 20 good: .$\displaystyle P(\text{0 bad}) \;=\;\frac{{5\choose0}{45\choose20}} {{50\choose20}}$
1 bad, 19 good: .$\displaystyle P(\text{1 bad}) \;=\;\frac{ {5\choose1}{45\choose19}} {{50\choose20}} $
. . Hence: .$\displaystyle P(\text{0 or 1 bad}) \;=\;\frac{{5\choose0}{45\choose20}} {{50\choose20}} + \frac{ {5\choose1}{45\choose19}} {{50\choose20}} $
Therefore: .$\displaystyle P(\text{at least 2 bad}) \;=\;1 - \left[\frac{{5\choose0}{45\choose20}} {{50\choose20}} + \frac{ {5\choose1}{45\choose19}} {{50\choose20}}\right] $
You can crank it out . . . I'll wait in the car.
.