# Math Help - Defective Widgets

1. ## Defective Widgets

In a certain factory, Machine A, Machine B, and Machine C are all producing widgets. Widgets produced by Machine A have a 1% chance of being defective. Likewise, widgets produced by Machine B and Machine C are defective 4% and 2% of the time, respectively. Of the total production of widgets in the factory, Machine A produces 30%, Machine B produces 25%, and Machine C produces 45%. Suppose a widget is selected at random from this factory.

a) What is the probability the widget is defective?
b) If the widget is defective, what is the probability it was produced by Machine B?

Any help would be appreciated. I believe that i am suppose to use the counting rule combination but am not sure because we are given two different sets of %...

2. Originally Posted by shannu82
In a certain factory, Machine A, Machine B, and Machine C are all producing widgets. Widgets produced by Machine A have a 1% chance of being defective. Likewise, widgets produced by Machine B and Machine C are defective 4% and 2% of the time, respectively. Of the total production of widgets in the factory, Machine A produces 30%, Machine B produces 25%, and Machine C produces 45%. Suppose a widget is selected at random from this factory.

a) What is the probability the widget is defective?
b) If the widget is defective, what is the probability it was produced by Machine B?

Any help would be appreciated. I believe that i am suppose to use the counting rule combination but am not sure because we are given two different sets of %...
for part a consider what is the prob that a widget is defective if it came from A

First the prob. that it came from A is .3 and the prob. that is is defective is .01

So the prob. that is came from A and is defective is

$(.3)(.01)=.003$

Now you need to repeat this for B and C and sum the probabilities. (why?)

For part b you need to use conditional probabilities

Let A be the even that the item is defective and B be the event that it came from machine B. Then

$P(B|A)=\frac{P(A \cap B)}{P(A)}$

Now use your result from part a.

3. Thank You for responding i just wanna make sure that i understand correctly...
a) P(A) =0.003 P(B) = 0.01 and P(C) = 0.009 Therefore the probability of defective widgets is 0.022.
b) using the formula above (0.003+0.01)/0.003 = 4.333% is the probability that the defective widget was created by Machine (B)?

4. Part A looks good..

For part B, TheEmptySet told you to assume A as the probability of the widgets being defective. This might have been ambiguous to you because you are using A as the probability of widget being made in factory A.

Suppose X is the probability that the widget is defective...

You already have P(X) from part(a) and you have $P(X \cap B)$ too..

Now find: $P(B|X)$