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Permutations, Combinations, Probability - Download Questions

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

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15 Dec 2007, 17:05
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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.

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15 Dec 2007, 22:20
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thanks a lot, they are very helpful.
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Re: P&C, Probability - Download [#permalink]

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25 Oct 2009, 22:26
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Expert's post
eresh wrote:
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.

Hope now it's clear.
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27 Dec 2007, 22:00
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thanks for the stuff, its really helpful.
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Re: Permutations, Combinations, Probability - Download Questions [#permalink]

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08 Dec 2010, 07:21
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mariyea wrote:
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

First of all:
 ! Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/Please post DS questions in the DS subforum: gmat-data-sufficiency-ds-141/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) = $$1-\frac{5}{16}=\frac{11}{16}$$

Answer: A.
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Re: P&C, Probability - Download [#permalink]

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03 Feb 2009, 23:20
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Thanks a lot, it is very helpfull!
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Re: P&C, Probability - Download [#permalink]

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14 Oct 2009, 06:21
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Thanks a lot friend, this can come in very handy.
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Re: Permutations, Combinations, Probability - Download Questions [#permalink]

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Re: P&C, Probability - Download [#permalink]

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19 May 2009, 10:34
Thanks...for the great work
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Re: P&C, Probability - Download [#permalink]

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24 Jun 2009, 04:22
This was really helpful. Thanks a lot dear!!, such inputs are of great help!!
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Re: P&C, Probability - Download [#permalink]

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24 Jun 2009, 12:10
This is great. What's the difference b/w the two files?
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Re: P&C, Probability - Download [#permalink]

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26 Jun 2009, 21:13
thx a lot.Looking for these for a long time.
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Re: P&C, Probability - Download [#permalink]

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27 Jun 2009, 05:07
Thanks for your help Amar
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Re: P&C, Probability - Download [#permalink]

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29 Jun 2009, 14:46
Thank you so much for your kindness
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Re: P&C, Probability - Download [#permalink]

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11 Jul 2009, 13:04
thanks a lot
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Re: P&C, Probability - Download [#permalink]

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27 Jul 2009, 22:57
Thaks a lot
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Re: P&C, Probability - Download [#permalink]

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28 Jul 2009, 04:18
thanks, i hope this is helpful.
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Re: P&C, Probability - Download [#permalink]

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29 Jul 2009, 11:58
To the moderator

Can we include the download to the "Combinations, Permutations, and Probability - References" Sticky post.

Thanks
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Re: P&C, Probability - Download [#permalink]

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29 Jul 2009, 18:31
I got more than half of the questions wrong in the first try
Seems I need a lot of practice.
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Re: P&C, Probability - Download [#permalink]

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19 Aug 2009, 10:21
Amardeep,

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!
Re: P&C, Probability - Download   [#permalink] 19 Aug 2009, 10:21

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