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

I have a question regarding the #37 question and the answer given.

37. How many 5 digit numbers can be created if the following terms apply: the leftmost digit is even, the second is odd, the third is a non even prime and the fourth and fifth are two random digits not used before in the number?

a) 2520 b) 3150 c) 3360 d) 6000 e) 7500

37. The best answer is A. The first digit has 4 options (2,4,6,8 and not 0), the second has 5 options (1,3,5,7,9) the third has 3 options (3,5,7 and not 2), the fourth has 7 options (10-3 used before) and the fifth has 6 options (10-4 used before). The total is 4*5*3*7*6=2520.

Now, it is not mentioned that a number cannot be repeated in for the 3rd digit. Say in the 2nd digit the number is 3 and so it is for the 3rd digit. The structure can be 233--, now for the 4th digit the option is 8 not 7.

Surely then the answer given is not right. But I don't know how to proceed further. Can someone please help? Or am I wrong about the porcess?

You are absolutely right. I'd say more there is no correct answer in choices. Surely the second and the third digits can be the same (in 3 cases out of 15) and in this case the number of possibilities for the 4th and 5th will be 8 and 7 and not 7 and 6.

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

Total=4*3*1*8*7+4*4*3*7*6=2688

P.S. eresh, can you please post other questions (you consider as worth of discussion) in PS or DS forums, as not many check this tree for particular problems, but rather for the tips and tutorials. You may even duplicate this question as there could be other opinions about it. Thanks.
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Re: Permutations, Combinations, Probability - Download Questions [#permalink]

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