# Defective Parts:

• May 29th 2011, 05:04 AM
hc215
Defective Parts:
Companies MPL, QS and RobotS supply machine parts in a 5:3:2 ratio respectively. 0.2% of the parts from MPL are defective. 0.1% are defective from QS and 0.5% are defective from RobotS.
For two machine parts chosen at random, calculate the probability that:
a) Both are supplied from MPL and both are defective.
b) Both supplied by one of the other companies and found to be defective.

(Being stupid I know, but blanked on this.)
Thank you.
• May 29th 2011, 02:55 PM
Soroban
Hello, hc215!

Quote:

Companies M, Q and R supply machine parts in a 5:3:2 ratio respectively.
0.2% of the parts from M are defective.
0.1% of the parts from Q are defective.
0.5% of the parts from R are defective.

We have the following data:

. . $\begin{array}{|c||c|c|}\hline \text{Company} & \text{Prob.} & P(\text{d{e}f}) \\ \hline\hline M & 0.5 & 0.002 \\ \hline Q & 0.3 & 0.001 \\ \hline R & 0.2 & 0.005 \\ \hline \end{array}$

Quote:

For two machine parts chosen at random, calculate the probability that:
a) Both are supplied from M and both are defective.

$P(\text{both M}\,\wedge\,\text{both d{e}f}) \;=\;(0.5)^2(0.002)^2 \;=\;0.000\,001$

Quote:

b) Both supplied by one of the other companies and both are defective.

$P(\text{Both Q}\,\wedge\,\text{both d{e}f}) \;=\;(0.3)^2(0.001)^2 \;=\;0.000\,000\,09$
$P(\text{Both R}\,\wedge\,\text{both d{e}f}) \;=\;(0.2)^2(0.005)^2 \;=\;0.000\,001$

$P\left([Q \vee R] \,\wedge\,\text{both d{e}f}\right) \;=\; 0.000\,000\,09 + 0.000\,001 \;=\;0.000\,001\,09$

• May 30th 2011, 02:00 AM
hc215
Thank you, Soroban.
Just wondering, does this method assume replacement of the parts? If so, how would you go about it without replacement? (I was getting confused with the second choice being conditional.)