Source: Official Guide Revised GRE 1st Ed. Part 8; Section 6; #24

2

Of the 20 lightbulbs in a box

Of the 20 lightbulbs in a box, 2 are defective. An inspector will select 2 lightbulbs simultaneously and at random from the box. What is the probability that neither of the lightbulbs selected will be defective? Give your answer as a fraction.

4 Explanations

1

Drasti Chaudhari

Why did you multiply and not add? Also, why is it not 1/18 and then 1/17? Isn't he picking just 1 bulb each time

Jun 13, 2018 • Comment

Sam Kinsman, Magoosh Tutor

Hi Drasti,

In probability, when we have two events that both need to happen, we have to multiply the probabilities of each event together. In other words, we need BOTH the first lightbulb to not be defective, AND the second lightbulb to not be defective. That means we have to multiply the two events together. You can read more about this rule here: https://magoosh.com/gmat/2012/gmat-math-probability-rules/

When the person picks the first lightbulb, the probability that it will not be defective is (# of non-defective bulbs) / (total # of bulbs). So we have: 18/20.

When the person picks the second lightbulb, there are only 19 lightbulbs left in the box. And, assuming that the first lightbulb that was picked was not defective, there are 17 nondefective lightbulbs in the box. So (# of non-defective bulbs) / (total # of bulbs) = 17/19.

I hope this helps! :)

Best,
Sam

Jun 16, 2018 • Reply

1

MIshal Patel

Hey Chris, which guide is this, because the one i have downloaded from your link doesn't have these questions.

Apr 8, 2016 • Comment

Cydney Seigerman, Magoosh Tutor

Hi there :) This is the 1st Edition of the Official Guide, which can be purchased in bookstores or sites such as amazon.com:

http://www.amazon.com/Official-Guide-revised-General-Test/dp/0071700528/ref=sr_1_1?ie=UTF8&qid=1318979846&sr=8-1

This question also appears in 2nd Edition of the OG (page 344, question 24).

Hope this helps :)

Apr 15, 2016 • Reply

2

mugen hassei

If I use the P(not a) rule(i.e probability of selecting a defective piece)-first try (2/20) probability that one gets a defective piece and 2nd try (1/19) is the probability that one gets a defective piece.Overall (2/20)(1/19)=1/190 is the probability that one gets a defective piece.Then probability of not selecting a defective piece is (1-1/190)=189/190.Where have I gone wrong?

Aug 7, 2015 • Comment

Cydney Seigerman, Magoosh Tutor

Hi Mugen :)

The solution you've proposed only considers the probability of choosing 2 defective bulbs. However, in order to use 1 - P(not A) = P(A), where A is choosing 2 working bulbs, we must consider all of the possible situations in which we would not choose 2 working bulbs, or in other words, situations in which we would choose at least 1 defective bulb. If we consider the selection in two stages, we can write out the possible ways of not selecting 2 working bulbs:

1. 1st: defective; 2nd: defective
2. 1st: defective; 2nd: working
3. 1st: working; 2nd: defective

As you can see, there are three different ways to not select two working bulbs. P(not A) is the sum of the probabilities that these sets of events occur. We can determine the probabilities of each set of events separately. I will use D for defective and W for working:

1. P(1.D 2.D) = 2/20*1/19 = 1/190
2. P(1.D 2.W) = 2/20*18/19 = 18/190
3. P(1.W 2.D) = 18/20*2/19 = 18/190

P(not A) = (1+18+18)/190 = 37/190.

Therefore, 1-P(not A) = 1 - 37/190 = 153/190.

This is the same answer that we would get if we only considered selecting 2 working bulbs. And as you can see, for this question, it is actually less complicated to find P(A) versus 1-P(not A). I hope this helps! :)

Sep 30, 2015 • Reply

3

Gravatar Chris Lele, Magoosh Tutor

Oct 11, 2012 • Comment

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