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Permutations, Combinations, Probability - Download Questions [#permalink]
15 Dec 2007, 17:05
68
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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.
Re: P&C, Probability - Download [#permalink]
25 Oct 2009, 22:26
3
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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.
Re: Permutations, Combinations, Probability - Download Questions [#permalink]
08 Dec 2010, 07:21
2
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Expert's post
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
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]
16 Feb 2014, 07:02
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Re: P&C, Probability - Download [#permalink]
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!
gmatclubot
Re: P&C, Probability - Download
[#permalink]
19 Aug 2009, 10:21
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