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saintforlife
The probability that a brand A widget is defective is x, the probability that a brand B widget is defective is 0.06, and the probability that a brand C widget is defective is 0.08. There are 40 brand A widgets, 35 brand B widgets, 25 brand C widgets, and nothing else in a container. If a widget is selected at random from the container, the probability that the widget selected is defective is 0.053. What is the value of x?

0.03
0.04
0.048
0.05
0.12
The expectation value of defectiveness is 100*0.053 = 5.3 which also is equal to (40x + 35*0.06 + 25* 0.08) = 40x + 2.1 + 2. Hence 40x=1.2, or x=0.03.
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Good qsn ; Option A for me !!
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Hi All,

Behind the "thick" wording (and slightly "scary" language) of this prompt, we're really just dealing with a "weighted average" question. The math approaches/calculations shown are spot-on, so I won't rehash any of that here. Instead, I want to show you how you can answer this question without doing much math at all. The specific numbers used in the prompt (and the ones listed in the answers) provide a 'concept shortcut' that you can use to your advantage. You'll still have to take notes and stay organized, but here's how to do it:

We're given info on 3 types of widgets: the number of each, and their respective likelihood of being defective:

Widget:
Brand A: 40 widgets, X% chance of defective
Brand B: 35 widgets, 6% chance of defective
Brand C: 25 widgets, 8% chance of defective

Total: 100 widgets, 5.3% chance of defective.

That 5.3% is interesting....since most of the widgets (35+25 = 60; more than half) have a GREATER % chance of being defective, the Brand A widgets must REALLY "bring down" the average.

If we focus on the Brand B and C widgets, we have 60 widgets and the average chance of defective is a little less than 7% (since there are more Brand B widgets, the weighted average would be slightly closer to 6% than 8%, thus < 7%).

So we have 60 widgets at just below 7% and we need to factor in 40 additional widgets to bring THAT average down to 5.3%....

IF....we had 60 Brand A widgets, then the chance of defective in A would be "balanced" around 5.3% (60 widgets near 7%, the 60 Brand A widgets near 3.5%). HOWEVER, we DON'T have 60 Brand A widgets - we have FAR FEWER - only 40 Brand A widgets that bring down the average. That means these Brand A widgets MUST have a chance of defective that is even LOWER than 3.5%. There's only one answer that fits.

Final Answer:
GMAT assassins aren't born, they're made,
Rich
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