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What is the probability that one of the two integers randomly selected from range 2029, inclusive, is prime and the other is a multiple of 3? (The numbers are selected independently of each other, i.e. they can be equal) A. 0.06 B. 0.12 C. 0.15 D. 0.18 E. 0.20
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16 Sep 2014, 01:16
Official Solution: What is the probability that one of the two integers randomly selected from range 2029, inclusive, is prime and the other is a multiple of 3? (The numbers are selected independently of each other, i.e. they can be equal) A. 0.06 B. 0.12 C. 0.15 D. 0.18 E. 0.20 Prime integers: 23 and 29. Multiples of 3: 21, 24, and 27. The probability that the first number is prime while the second is a multiple of \(3 = \frac{2}{10} \frac{3}{10} = 0.06\). The probability that the first number is a multiple of 3 while the second is prime \(= \frac{3}{10} \frac{2}{10} = 0.06\). The probability that one of the two integers is prime and the other is a multiple of \(3 = 0.06 + 0.06 = 0.12\). Answer: B
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Re: M2208
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03 Apr 2016, 06:15
Hi Bunuel I was wondering why the order of the selection matters here ? Thanks Bunuel wrote: Official Solution:
What is the probability that one of the two integers randomly selected from range 2029 is prime and the other is a multiple of 3? (The numbers are selected independently of each other, i.e. they can be equal)
A. 0.06 B. 0.12 C. 0.15 D. 0.18 E. 0.20
Prime integers: 23 and 29. Multiples of 3: 21, 24, and 27. The probability that the first number is prime while the second is a multiple of \(3 = \frac{2}{10} \frac{3}{10} = 0.06\). The probability that the first number is a multiple of 3 while the second is prime \(= \frac{3}{10} \frac{2}{10} = 0.06\). The probability that one of the two integers is prime and the other is a multiple of \(3 = 0.06 + 0.06 = 0.12\).
Answer: B



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Re: M2208
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04 Apr 2016, 06:38
Michael KC Chen wrote: Hi Bunuel I was wondering why the order of the selection matters here ? Thanks Bunuel wrote: Official Solution:
What is the probability that one of the two integers randomly selected from range 2029 is prime and the other is a multiple of 3? (The numbers are selected independently of each other, i.e. they can be equal)
A. 0.06 B. 0.12 C. 0.15 D. 0.18 E. 0.20
Prime integers: 23 and 29. Multiples of 3: 21, 24, and 27. The probability that the first number is prime while the second is a multiple of \(3 = \frac{2}{10} \frac{3}{10} = 0.06\). The probability that the first number is a multiple of 3 while the second is prime \(= \frac{3}{10} \frac{2}{10} = 0.06\). The probability that one of the two integers is prime and the other is a multiple of \(3 = 0.06 + 0.06 = 0.12\).
Answer: B Hi, more than often follow this rule if in doubt..1) If you are picking two simultaneously/together, you do not have any order in place.. 2) whenever you are picking two with/ without repetition, it can be picked as either A and B or B and A..
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Re: M2208
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04 Apr 2016, 08:11
chetan2u wrote: Michael KC Chen wrote: Hi Bunuel I was wondering why the order of the selection matters here ? Thanks Bunuel wrote: Official Solution:
What is the probability that one of the two integers randomly selected from range 2029 is prime and the other is a multiple of 3? (The numbers are selected independently of each other, i.e. they can be equal)
A. 0.06 B. 0.12 C. 0.15 D. 0.18 E. 0.20
Prime integers: 23 and 29. Multiples of 3: 21, 24, and 27. The probability that the first number is prime while the second is a multiple of \(3 = \frac{2}{10} \frac{3}{10} = 0.06\). The probability that the first number is a multiple of 3 while the second is prime \(= \frac{3}{10} \frac{2}{10} = 0.06\). The probability that one of the two integers is prime and the other is a multiple of \(3 = 0.06 + 0.06 = 0.12\).
Answer: B Hi, more than often follow this rule if in doubt..1) If you are picking two simultaneously/together, you do not have any order in place.. 2) whenever you are picking two with/ without repetition, it can be picked as either A and B or B and A.. what is wrong when I do: (2C1 x 3C1)/(10C2). Please explain.



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Re: M2208
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04 Apr 2016, 08:24
robu wrote: chetan2u wrote:
what is wrong when I do: (2C1 x 3C1)/(10C2). Please explain.
Hi , two errors1) 10C2 is wrong because it is given that the same number can be picked up again..In other words it is with REPLACEMENT, that is number will remain 10 each time.. so total ways will be 10*10 2) 2C1 * 3C2..there are two ways here 2C1*3C1 OR 3C1*2C1 so two ways 2C1*3C1*2
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Re: M2208
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07 Sep 2016, 00:00
chetan2u wrote: Michael KC Chen wrote: Hi Bunuel I was wondering why the order of the selection matters here ? Thanks Bunuel wrote: Official Solution:
What is the probability that one of the two integers randomly selected from range 2029 is prime and the other is a multiple of 3? (The numbers are selected independently of each other, i.e. they can be equal)
A. 0.06 B. 0.12 C. 0.15 D. 0.18 E. 0.20
Prime integers: 23 and 29. Multiples of 3: 21, 24, and 27. The probability that the first number is prime while the second is a multiple of \(3 = \frac{2}{10} \frac{3}{10} = 0.06\). The probability that the first number is a multiple of 3 while the second is prime \(= \frac{3}{10} \frac{2}{10} = 0.06\). The probability that one of the two integers is prime and the other is a multiple of \(3 = 0.06 + 0.06 = 0.12\).
Answer: B Hi, more than often follow this rule if in doubt..1) If you are picking two simultaneously/together, you do not have any order in place.. 2) whenever you are picking two with/ without repetition, it can be picked as either A and B or B and A.. It's not actually like that. Check this thread ( probabilityofsimultaneouseventsveritasvsmgat10599420.html) Picking simultaneously is the same as picking one by one without replacement. In any case you need to multiply by the possible number of combinations. The only case, when you don't need to multiply is when the order is strictly set  for instance, what's the probability that the FIRST number will be prime and the second will be a multiple of 3.



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Re: M2208
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27 May 2017, 07:36
Can we get a confirmation that whether 0.06 is the answer or 0.12?
I am not sure whether we have to count the probability twice  i.e. for each case.
@Bunnel  can you please confirm.



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Re: M2208
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27 May 2017, 08:49
Vikram_Katti wrote: Can we get a confirmation that whether 0.06 is the answer or 0.12?
I am not sure whether we have to count the probability twice  i.e. for each case.
@Bunnel  can you please confirm. The correct answer is B  0.12
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Re: M2208
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10 Jul 2017, 08:37
I don't agree with the answer  if order is matter then it matter for both number of desirable events and for total number of events. SO total number of events is not 10x10 but 10x10x2 and desirable events are 3x4x2 also ! Then the answer is 0.06
The reply should be wrong here



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Re: M2208
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10 Jul 2017, 11:50
alexxko wrote: I don't agree with the answer  if order is matter then it matter for both number of desirable events and for total number of events. SO total number of events is not 10x10 but 10x10x2 and desirable events are 3x4x2 also ! Then the answer is 0.06
The reply should be wrong here This is not right. The order does matter  (prime, multiple of 3) and (multiple of 3, prime) are tow different cases. The reasoning as to why it should be 10*10*2 is not correct and not clear. For both cases we are choosing from 10 numbers, that's why it is 2/10*3/10 + 3/10*2/10.
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Re: M2208
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11 Jul 2017, 00:44
Bunuel wrote: What is the probability that one of the two integers randomly selected from range 2029 is prime and the other is a multiple of 3? (The numbers are selected independently of each other, i.e. they can be equal)
A. 0.06 B. 0.12 C. 0.15 D. 0.18 E. 0.20 Dear Bunuel, I rarely post about phrasing your question but if you add word 'inclusive' after range 2029, it would be clear. Thanks



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Re: M2208
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11 Jul 2017, 02:29
Mo2men wrote: Bunuel wrote: What is the probability that one of the two integers randomly selected from range 2029 is prime and the other is a multiple of 3? (The numbers are selected independently of each other, i.e. they can be equal)
A. 0.06 B. 0.12 C. 0.15 D. 0.18 E. 0.20 Dear Bunuel, I rarely post about phrasing your question but if you add word 'inclusive' after range 2029, it would be clear. Thanks Edited as suggested. Thank you.
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Re: M2208
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03 Jan 2018, 08:58
chetan2u wrote: robu wrote: chetan2u wrote:
what is wrong when I do: (2C1 x 3C1)/(10C2). Please explain.
Hi , two errors1) 10C2 is wrong because it is given that the same number can be picked up again..In other words it is with REPLACEMENT, that is number will remain 10 each time.. so total ways will be 10*10 2) 2C1 * 3C2..there are two ways here 2C1*3C1 OR 3C1*2C1 so two ways 2C1*3C1*2 Thanks for the explanation !!!
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Re: M2208
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04 Jan 2018, 03:55
What must change in the question stem, in order the result to be 0.06? This is, no need to sum both probabilities. Please clarify! Thanks



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Re: M2208
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31 Jan 2018, 00:14
Hello Bunuel: Bunuel wrote: alexxko wrote: I don't agree with the answer  if order is matter then it matter for both number of desirable events and for total number of events. SO total number of events is not 10x10 but 10x10x2 and desirable events are 3x4x2 also ! Then the answer is 0.06
The reply should be wrong here This is not right. The order does matter  (prime, multiple of 3) and (multiple of 3, prime) are tow different cases. The reasoning as to why it should be 10*10*2 is not correct and not clear. For both cases we are choosing from 10 numbers, that's why it is 2/10*3/10 + 3/10*2/10. Can you please explain why does order matter here? In other words, how is 23, 27 different from 27, 23 as in both cases it's the same prime number and the same multiple of 3 that's being counted. Thanks!



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Re: M2208
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31 Jan 2018, 02:00
sanjay1810 wrote: Hello Bunuel: Bunuel wrote: alexxko wrote: I don't agree with the answer  if order is matter then it matter for both number of desirable events and for total number of events. SO total number of events is not 10x10 but 10x10x2 and desirable events are 3x4x2 also ! Then the answer is 0.06
The reply should be wrong here This is not right. The order does matter  (prime, multiple of 3) and (multiple of 3, prime) are tow different cases. The reasoning as to why it should be 10*10*2 is not correct and not clear. For both cases we are choosing from 10 numbers, that's why it is 2/10*3/10 + 3/10*2/10. Can you please explain why does order matter here? In other words, how is 23, 27 different from 27, 23 as in both cases it's the same prime number and the same multiple of 3 that's being counted. Thanks! The point is that those are two different events. Imagine that there are 10 balls with integer values from 20 through 29 on them. We are picking two balls with replacement. The case when the first ball has a prime number on it and the second ball has a multiple of 3 on it is a different event from a case when the first ball has a multiple of 3 on it and the second ball has a prime number on it.
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25 Mar 2019, 12:26
I arrived at
(\(\frac{2}{10}\)*\(\frac{3}{10}\))*2 and everything seemed so easy with this 95% hard question ... BUT
The Ozone layer became thinner, The moon showed the dark side and my brain decided that 2/10=1/2
Ended up  Wrong
I coursed math **** this sh@t  I shouted throwing papers away ... Several seconds later: I have to pass  gathering my goddamn papers again
Peace of mind, Peace of mind, Peace of mind










