MHB Mechanical Part Defects in Lot of 500 Pieces

[paulmdrdo1]

A certain mechanical part of a machine can be defective because it has one or more out of
three possible defects: insufficient tensile strength, a burr or a diameter outside of tolerance
limits. In a lot of 500 pieces
19 have tensile strength defect,
17 have a burr,
11 have an unacceptable diameter,
12 have tensile strength and burr defects,
7 have tensile strength and diameter defects,
5 have burr and diameter defects,
2 have all three defects.

I have tried creating venn diagram to sove this but I end having a convoluted solution. Can you help go about using a more managable method? Regards!

 

[Jameson]

A certain mechanical part of a machine can be defective because it has one or more out of
three possible defects: insufficient tensile strength, a burr or a diameter outside of tolerance
limits. In a lot of 500 pieces
19 have tensile strength defect,
17 have a burr,
11 have an unacceptable diameter,
12 have tensile strength and burr defects,
7 have tensile strength and diameter defects,
5 have burr and diameter defects,
2 have all three defects.

I have tried creating venn diagram to sove this but I end having a convoluted solution. Can you help go about using a more managable method? Regards!

Hi paulmdrdo,

You want to draw a venn diagram like this:

Notice that there are 7 regions in this diagram. The trick to solving these problems is to start in the middle and work your way out. Otherwise the double and triple counting can be quite the headache. So we know the number with all three defects is 2. That two goes in the middle. Now you can label each circle as you wish. Call one tensile strength defect, another burr, and the last one unacceptable diameter. It doesn't matter which way you label them.

We know that 5 have burr and diameter defects. If they have burr and diameter, then the group in the middle counts right since they have all three right? So 5-2=3 have JUST burr and diameter defects. Lastly, 17 total have burr. We've already accounted for those with all three defects and two of them. So that leaves 17-2-3=12 with JUST burr.

You repeat the process for the other two, starting in the middle. It won't add up to 500 but that makes sense in the context of the problem since these are defects.

 

[Evgeny.Makarov]

You can also use the inclusion-exclusion principle, which avoids geometrical intuition and makes finding the total number of defective parts very straightforward. Namely,
\[
|A_1\cup A_2\cup A_3|=|A_1|+|A_2|+|A_3|-|A_1\cap A_2|-|A_1\cap A_3|-|A_2\cap A_3|+|A_1\cap A_2\cap A_3|.
\]
Here $A_1$, $A_2$, $A_3$ are sets, $|A|$ is the number of elements in $A$, $A\cap B$ is the intersection of $A$ and $B$ and $A\cup B$ is the union of $A$ and $B$.