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Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]
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21 Sep 2015, 12:53
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alexeykaplin wrote: I'm little bit confused here. The way provided is pretty logical and I like it but I have a conundrum with more choosey approach.
In Set A we have 12 integers, these ones divisble by 3  12, 15,18, 21  4 numbers In set B we have 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47  11 prime numbers
prob of xy divisible by 3 = (xy's divisble by 3)/(total possible xy's)
xy's divisible by 3= 11x4=44 total possible xy's= 11x124=128, I deduct 4 to avoid double counting of 11, 13, 17,19 as these appear in both sets
so, I have 44/128=11/32, slightly more that 1/3.
please, tell me where am I wrong here. It is already given that y is being chosen from the list of prime numbers, and hence there will be a probability of 1 if we choose any number from y. And for x, we have to find the probability of choosing a multiple of 3. Hence, the final probability is 4/12*1 = 1/3
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Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]
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21 Sep 2015, 13:34
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tarunktuteja wrote: y has no factor z such that 1 < z < y this means that y is a prime and it has to lie between 10 and 50 inclusive. Now, since the product of xy is divisible by 3 so that means that we have to look for multiples of 3 in set A since primes from set B are all greater than 3. Set B has 12 elements and out of which 4 are multiples of 3 so probability should be 4/12 = 1/3.
Consider giving kudos if my post helped. After following all the discussion on this thread, I feel that considering only x is not correct. The question clearly states that y is the probability of selected a prime number from the set B. And then it asks for the probability of product xy being divisible by 3! Clearly we cannot assume that y will always be prime. We need to take it into consideration that we have picked y as a prime number from B and x as a factor of 3 from set A. The events x and y are definitely not mutually exclusive in this case as we need to find the probability for product xy . Also there can be cases where y if not prime could be a multiple of 3 and then xy would still be divisible by 3! Please highlight if I'm overlooking something, But clearly the options and OA isnt correct.



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Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]
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21 Sep 2015, 14:04
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itwarriorkarve wrote: tarunktuteja wrote: y has no factor z such that 1 < z < y this means that y is a prime and it has to lie between 10 and 50 inclusive. Now, since the product of xy is divisible by 3 so that means that we have to look for multiples of 3 in set A since primes from set B are all greater than 3. Set B has 12 elements and out of which 4 are multiples of 3 so probability should be 4/12 = 1/3.
Consider giving kudos if my post helped. After following all the discussion on this thread, I feel that considering only x is not correct. The question clearly states that y is the probability of selected a prime number from the set B. And then it asks for the probability of product xy being divisible by 3! Clearly we cannot assume that y will always be prime. We need to take it into consideration that we have picked y as a prime number from B and x as a factor of 3 from set A. The events x and y are definitely not mutually exclusive in this case as we need to find the probability for product xy . Also there can be cases where y if not prime could be a multiple of 3 and then xy would still be divisible by 3! Please highlight if I'm overlooking something, But clearly the options and OA isnt correct. "y is a number chosen randomly from Set B, and y has no factor z such that 1 < z < y, what is the probability that the product xy is divisible by 3" you are missing the bold face part. It is already given that we are choosing y among the set which is govern by 1<z<y. If this condition was not given, we would have followed your approach.
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Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]
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29 Jun 2016, 12:41
I think this is a bad question.. The way its written it seems that you have to chose 4 out of 11 (the multiples of 3 from 10 to 21) and then it will have to match a prime chosen from 10 to 50, so 11 possibilities of out 50. So we get 4/11*11/40.
The question should read that a number WAS CHOSEN from bla bla, so its clear that the action of choosing it from the respective sample has already taken place.
Anyways maybe its just me..



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Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]
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29 Jun 2016, 12:58
and y has no factor z such that 1 < z < y Question poorly worded, does the probability includes the drawing of y ?



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Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]
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11 Mar 2017, 12:12
If y has no factor z such that 1 < z < y, then y must be prime. Let's look at a few examples to see why this is true: 6 has a factor 2 such that 1 < 2 < 6: 6 is NOT prime 15 has a factor 5 such that 1 < 5 < 15: 15 is NOT prime 3 has NO factor between 1 and 3: 3 IS prime 7 has NO factor between 1 and 7: 7 IS prime Because it is selected from Set B, y is a prime number between 10 and 50, inclusive. The only prime number that is divisible by 3 is 3, so y is definitely not divisible by 3. Thus, xy is only divisible by 3 if x itself is divisible by 3. We can rephrase the question: “What is the probability that a multiple of 3 will be chosen randomly from Set A?” There are 21 – 10 + 1 = 12 terms in Set A. Of these, 4 terms (12, 15, 18, and 21) are divisible by 3. There are 21 – 10 + 1 = 12 terms in Set A. Of these, 4 terms (12, 15, 18, and 21) are divisible by 3. Thus, the probability that x is divisible by 3 is : 4/12 = 1/3
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Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]
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12 Mar 2017, 04:24
Though the question is simple, so much prose makes it seem complicated. 'y' has to be a prime number as the integer randomly selected cannot have a factor 'z' of the form 1<z<y.
y=11,13,17,19,23,29,31,37,41,43 (Seeing this sequence it can be determined that none of the values are divisible by 3)
x=10,11,12,13,14,15,16,17,18,19,20,21 (12,15,18,21 are the only integers divisible by 3). Hence probability = 4/12 = 1/3



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Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]
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20 Mar 2018, 06:02
daviesj wrote: Set A consists of all the integers between 10 and 21, inclusive. Set B consists of all the integers between 10 and 50, inclusive. If x is a number chosen randomly from Set A, y is a number chosen randomly from Set B, and y has no factor z such that 1 < z < y, what is the probability that the product xy is divisible by 3?
A. 1/4 B. 1/3 C. 1/2 D. 2/3 E. 3/4 Set y that satisfies the given criteria : { 11,13,17,19,23,29,31,37,41,43,47,} set x = { 12,15,18,21} No. of pairs of XY that are divisible by 3 = 44 Total xy = total 12 elements of set x * 11 elements of set y that satisfy the condition , hence total xy = 12 *11 Probability = \(\frac{44}{12*11}\) = \(\frac{1}{3}\) Main point of confusion may be when selecting the total number of pairs , remember when selecting one element of Y we have a certain condition that it should be prime , when selecting one element from x we have no such condition. Hence total cases are 12 * 11 .Out of which favorable cases are 4*11 = 44. Hence Probability = \(\frac{44}{12*11}\) = \(\frac{1}{3}\) Hope this helps .
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Re: Set A consists of all the integers between 10 and 21, inclus [#permalink]
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20 Mar 2018, 10:08
daviesj wrote: Set A consists of all the integers between 10 and 21, inclusive. Set B consists of all the integers between 10 and 50, inclusive. If x is a number chosen randomly from Set A, y is a number chosen randomly from Set B, and y has no factor z such that 1 < z < y, what is the probability that the product xy is divisible by 3?
A. 1/4 B. 1/3 C. 1/2 D. 2/3 E. 3/4 y has no factor z such that 1 < z < y, implied that y is a prime number. Therefore x must be a multiple of 3 and is between 10 and 21 inclusive, we got the following number: 12,15,18,21 The probability that xy is divisible by 3: \(\frac{4}{12}\) = \(\frac{1}{3}\) => Answer (B)




Re: Set A consists of all the integers between 10 and 21, inclus
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