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Permutations, Combinations, Probability - Download Questions [#permalink]

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15 Dec 2007, 18:05

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Guys,

Enclosed please find list of 55 different question with explanation. I got this from the GMATClub forum and found very impressive to learn most of the required commonly tested concepts.

I am returning back to the forum so that everyone can use it for hir/her benefit.

Amar

Please discuss specific questions in PS or DS subforums.

Bunuel, I believe your solution is absolutely correct.

So here is my solution: 1. second and third digits are the same: 4*3*1*8*7

2. second and third digits are not the same: 4*4*3*7*6

Thinking of the approach, I understand that there are a total 15 combination of the 2nd and 3rd digit, of which in 3 combination the digits will be same. Therefore, there are 12 other possibilities where the digits are not same. The red part 4*3 also is equal to 12 but I could not figure out why it was 4*3 (My brains seems to stop working when I try to focus too much :D). Anyway, your answer seems correct and thank you for that.

Think about it this way: we don't want second and third digits to be the same, let's first choose the third digit(clearly it doesn't matter which one we choose first) how many possibilities are there? 3 (3 odd primes), than choose the second digit how many possibilities are there as one odd prime is already used? 5-1=4 --> 3*4=4*3.

25. John wrote a phone number on a note that was later lost. John can remember that the number had 7 digits, the digit 1 appeared in the last three places and 0 did not appear at all. What is the probability that the phone number contains at least two prime digits?

a) 15/16 b) 11/16 c) 11/12 d) ½ e) 5/8

I don't quite understand the explanation to this problem. Can someone please help me out? Thanks in advance

No posting of PS/DS questions is allowed in the main Math forum.

As for the question:

The phone numbers is of a type: {X}{X}{X}{X}{1}{1}{1}.

{X}'s can take following values: 4 primes {2, 3, 5, 7} and 4 non-primes {4, 6, 8, 9}. Total 8 choices for each {X}. Probability that {X} will be prime is therefore \(\frac{4}{8}=\frac{1}{2}\) and probability of {X} will not be a prime is again \(\frac{1}{2}\).

We want at least 2 {X}'s out of 4 to be primes, which means 2, 3 or 4 primes.

Let's count the opposite probability and subtract it from 1.

Opposite probability of at least 2 primes is 0 or 1 prime:

So {P}{NP}{NP}{NP} and {NP}{NP}{NP}{NP}.

Scenario 1 prime - {P}{NP}{NP}{NP}: \(\frac{4!}{3!}*\frac{1}{2}*(\frac{1}{2})^3=\frac{4}{16}\). We are multiplying by \(\frac{4!}{3!}\) as scenario {P}{NP}{NP}{NP} can occur in several different ways: {P}{NP}{NP}{NP}, {NP}{P}{NP}{NP}, {NP}{NP}{P}{NP}, {NP}{NP}{NP}{P} - 4 ways (basically the # of permutations of 4 objects out ow which 3 are the same).

Scenario 0 prime - {NP}{NP}{NP}{NP}: \((\frac{1}{2})^4=\frac{1}{16}\).

Hence opposite probability = \(\frac{4}{16}+\frac{1}{16}=\frac{5}{16}\).

So probability of at least 2 primes is: 1-(Opposite probability) = \(1-\frac{5}{16}=\frac{11}{16}\)

Re: Permutations, Combinations, Probability - Download Questions [#permalink]

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16 Feb 2014, 08:02

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Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Thank you so much for your very helpful post. I was in the middle of compiling a list of probablity questions myself when I realized that you had already done so.

Thanks again!

gmatclubot

Re: P&C, Probability - Download
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19 Aug 2009, 11:21

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