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by Taiana Melo » Tue Mar 29, 2016 4:52 pm
Ìf a quality control check is made by inspecting a sample of 2 lightbulbs from a box of 12 lightbulbs, how many different samples can be chosen?
6
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by [email protected] » Tue Mar 29, 2016 9:32 pm
Hi Taiana Melo,

This looks like a PS question (not an IR question), so you should have posted it here:

https://www.beatthegmat.com/problem-solving-f6.html

That having been said, I'll be happy to walk you through it. This is an example of a straight-forward Combinations question. When dealing with Combinations, it often helps to use the Combination Formula:

N! / K!(N-K)! where N is the total number of items and K is the number in the 'sub-group'

Here, we have 12 lightbulbs and we're choosing groups of 2. Since it doesn't matter which lightbulb comes first (choosing bulb A and then bulb B is the SAME as choosing bulb B and then bulb A), the order of the bulbs does NOT matter. THAT is a big 'clue' to use the Combination Formula:

12! / 2!(12-2)! = 12! / 2!10! = (12)(11) / (2)(1) = 132/2 = 66

Final Answer: B

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by MartyMurray » Tue Mar 29, 2016 9:40 pm
You are selecting two lightbulbs out of 12.

The first bulb is any one of 12.

The second bulb is one of the remaining 11.

So there are 12 choices for the first bulb and for every 12 there are 11 choices for the second bulb. Therefore your next step is to multiply 12 x 11 to get 132.

However, 132 is not the correct answer as it contains samples picked twice in opposite ways.

For instance, you could have picked bulb 1 first and bulb 2 second.

Alternatively, you could have picked bulb 2 first and bulb 1 second.

Those two sets of choices would produce the same sample. So you need to account for the fact that the same two bulbs can be chosen in two different orders.

That is done by dividing (12 x 11)/2! = 132/2 = 66.

The correct answer is D.
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