Source: Official Guide Revised GRE 1st Ed. Part 6; Set 2; #10

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Alyce Kayes

Would it be okay to use the Combination formula here? I still managed to get P(neither are lawyers) = 0.68... which rounds to the correct answer 0.7 (C).

I set it up as (580 choose 2)/(700 choose 2) since the numerator is # of ways 2 lawyers can be chosen and the denominator is total # of ways to choose 2 people.

I will say that this calculation was not super fast to do, and I did use a calculator. Perhaps this method works but isn't recommended because of how long it would take? Thanks!

Sep 22, 2017 • Comment

Sam Kinsman, Magoosh Tutor

Hi Alyce! Yes, combinations work here! However, as you pointed out, its a really long process to get to the answer. 580 choose 2 = 580! / (2! * 578!) = 167,910. That's a really big number, and getting there will take a lot of steps.

Oct 19, 2017 • Reply

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Mridula Polina

Can we do a 1-(120/700)x(120/700) here?

Nov 9, 2015 • Comment

Pete Rossman, Magoosh Tutor

Hi Mridula! I see what you're trying to do by taking the complement that both people chosen are lawyers. However, what's remaining isn't the probability that none are lawyers: it's the probability that both chosen are NOT lawyers, so there's still a possibility that we have one lawyer chosen as a member. As such, we can't take the complement that both are lawyers for this problem.

Nov 14, 2015 • Reply

Oscar Nnoli

This explanation is confusing, could you please elaborate. The statement made in the second sentence is a bit hard to comprehend.

Feb 15, 2016 • Reply

Cydney Seigerman, Magoosh Tutor

Hi Oscar :)

The complement rule states

P(not A) = 1 - P(A)

(120/700)x(120/700) is the probability that both people chosen are lawyers. So, if we apply the complement rule using this probability, we will have: the quantity

1 - P(both lawyers)

As Pete points out, the difference 1 - P(both lawyers) doesn't equal P(no lawyers) but rather P(1 lawyer or no lawyers). This is because in this problem "not A" (or not choosing both lawyers) can be achieved in two ways:

1. no lawyers

2. 1/2 lawyers

For that reason, as Pete notes, we're not able to use the solution Mridula proposed to solve for probability that no lawyers are selected in this problem.

I hope this clears things up! :)

Feb 15, 2016 • Reply

Cydney Seigerman, Magoosh Tutor

You're very welcome, Oscar! We're happy to have helped make this more clear :D

Feb 19, 2016 • Reply

Simardeep Singh

Can't we do (1- (120/700)*(119/699)). Coz for the first case, the lawyer will be choosen. And for the second case, again the lawyer will br choosen.

Jul 23, 2016 • Reply

Cydney Seigerman, Magoosh Tutor

Hi Simardeep :)

You're correct that using your set-up, we would calculate the 2 lawyers would be chosen. However, and as we mention above, the complement of "no lawyers chosen" is NOT "both are lawyers" but rather "at least 1 lawyer." And there are three ways to have "at least 1 lawyer":

1. Both are lawyers (the probability in your set-up)

2. First one picked is a lawyer, but the second isn't.

3. Second one picked is a lawyer, but the first isn't.

So, to use the complement rule, we would need to calculate the probability of all three situations and subtract them from 1.

Hope this clears up your doubts! :)

Jul 24, 2016 • Reply

Rachit Singh

I understood the explanation quite well for this solution. However, just out of curiosity, I had tried solving this question from the complement rule which you had just mentioned above. I have doubt in this steo: after finding those three individual probabilities, do we need to add them and then subtract from 1 OR multiply those 3 probabilities and subtract from 1?

Also, let me know if the individual probabilities are correct.

Both are lawyers- (120/700)x(119x699)

First is lawyer, but second- (120/700)x(580/699)

Second is not a lawyer, but first is a lawyer- (580/700)x(120/699)

Sep 5, 2016 • Reply

Cydney Seigerman, Magoosh Tutor

Hi Rajendra!

Happy to clarify :) We have to add the individual probabilities together and subtract this sum from 1. That's because the 3 situations are mutually exclusive (they cannot occur at the same time). Also, your set up to find the three probabilities looks good! :D

Sep 5, 2016 • Reply

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Official Guide Revised GRE 1st Ed.

Official Guide Revised GRE 2nd Ed.

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