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

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19 May 2010, 05:54

Well here is another big thank you!
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Re: Permutations, Combinations, Probability - Download Questions [#permalink]

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08 Dec 2010, 07:15

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

I don't quite understand the explanations to these problems. Can someone please help me out? Thanks in advance
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Last edited by mariyea on 08 Dec 2010, 07:52, edited 1 time in total.

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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08 Dec 2010, 08:18

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

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

Answer: A.

This is just great! Thanks Bunuel!!!
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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.

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

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06 Jan 2011, 04:36

Thanks for the upload mate!

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?

gmatclubot

Re: Permutations, Combinations, Probability - Download Questions
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06 Jan 2011, 04:36

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