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19 May 2010, 06:54
Well here is another big thank you!
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22 May 2010, 18:03
this is just what i needed!
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21 Jun 2010, 17:08
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23 Jun 2010, 20:01
very very useful. thanks for this man!
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24 Jun 2010, 11:48
Thanks a lot. I 'm having a hard time with permutation and combination
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25 Jun 2010, 04:34
Great stuff, Thanks !
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21 Jul 2010, 06:17
Thanks for the nice stuff.
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01 Sep 2010, 04:58
thanks )))
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05 Sep 2010, 02:22
Thanx a lot dear!!but i found no diff in two files!!very useful for me!!
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14 Sep 2010, 08:42
Thank you so much. I really need practice in probability and combinations. This will help so much.
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02 Dec 2010, 17:39
My weakest math area. Fortunate to see this file.
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06 Dec 2010, 08:09
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08 Dec 2010, 08:15
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

29. In a jar there are 15 white balls, 25 red balls, 10 blue balls and 20 green balls. How many balls must be taken out in order to make sure we took out 8 of the same color?

a) 8
b) 23
c) 29
d) 32
e) 53

30. In a jar there are 21 white balls, 24 green balls and 32 blue balls. How many balls must be taken out in order to make sure we have 23 balls of the same color?

a) 23
b) 46
c) 57
d) 66
e) 67

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Last edited by mariyea on 08 Dec 2010, 08:52, edited 1 time in total.
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08 Dec 2010, 08:21
2
KUDOS
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

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) = $$1-\frac{5}{16}=\frac{11}{16}$$

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08 Dec 2010, 09:18
Bunuel wrote:
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

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) = $$1-\frac{5}{16}=\frac{11}{16}$$

This is just great! Thanks Bunuel!!!
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25 Dec 2010, 10:05
Hey! Thanks a lot for the effort! What's the source of these questions? Any of them from GMATPrep or Kaplan/MGMAT practice tests???
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25 Dec 2010, 11:22
I'm looking forward to use these questions to hone my Probability & Combinatorics skills.
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03 Jan 2011, 20:23
Shouldn't "zero" counted as even number in the leftmost digit or am I wrong?

So, there should be 5

Bunuel wrote:
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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03 Jan 2011, 21:00
I saw my mistake and I fold...

I thought this was one of the credit-card numbers type of question..
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06 Jan 2011, 05:36

Quick question on number 4 though:

How many 3-digit numbers satisfy the following conditions: The first digit is different from zero and the other digits are all different from each other?

From my reading of this "the other digits" implies the 2nd and 3rd digits and not the first. Well if only the 2nd and 3rd digits are different from each other then shouldn't it be 9*10*9?

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