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If two of the four expressions x+y, x+5y, xy, and 5xy are chosen at [#permalink]
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15 Apr 2010, 04:19
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If two of the four expressions x+y, x+5y, xy, and 5xy are chosen at random, what is the probability that their product will be of the form of x^2 (by)^2, where b is an integer? A. 1/2 B. 1/3 C. 1/4 D. 1/5 E. 1/6
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If two of the four expressions x+y, x+5y, xy, and 5xy are chosen at [#permalink]
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15 Apr 2010, 04:41




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Re: If two of the four expressions x+y, x+5y, xy, and 5xy are chosen at [#permalink]
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12 Nov 2010, 12:12
it should be 1/2 * 1/3 = 1/6
First case you can chose either x+y or xy. So you have 2 options out of 4. second time you have 1 option out of 3




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Re: If two of the four expressions x+y, x+5y, xy, and 5xy are chosen at [#permalink]
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21 Sep 2010, 22:54
vigneshpandi wrote: Can anyone explain how is this derived? Attachment: Probability of the product.JPG The only way to get x^2  (by)^2 is if you multiply (x+by) with (xby). Looking at the expressions given, the only pair that will satisfy this is (x+y) & (xy) No of ways to pick pairs = C(4,2) = 6 No of pairs satisfying condition = 1 Probability = 1/6
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Re: If two of the four expressions x+y, x+5y, xy, and 5xy are chosen at [#permalink]
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12 Nov 2010, 11:48
Hey guys, I know its a stupid question, but for some reason i cannot figure it out now... Maybe Im just too tired... Why cant we do this questions simpler? as in  1/4*1/3? I understand perfectly the 4C2 way, but i want to understand the others as well... Thanks. Sry for the stupidity...
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Re: If two of the four expressions x+y, x+5y, xy, and 5xy are chosen at [#permalink]
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24 Apr 2012, 02:04
How do I know from the phrasing of this problem that, once chosen, an expression is taken off the list (ie not replaced)? When I was solving the problem I assumed that an expression of form (x+y)(x+y) is possible. In that case the number of combinations is 8.



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If two of the four expressions x+y, x+5y, xy, and 5xy are chosen at [#permalink]
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24 Apr 2012, 12:52
alexcey wrote: How do I know from the phrasing of this problem that, once chosen, an expression is taken off the list (ie not replaced)? When I was solving the problem I assumed that an expression of form (x+y)(x+y) is possible. In that case the number of combinations is 8. Question says: "If two of the four expressions x+y, x+5y, xy, and 5xy are chosen at random..." so, at least for me, it's quite obvious that there is no replacement whatsoever. Also if it were the case the question would explicitly mention that.
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Re: If two of the four expressions x+y, x+5y, xy, and 5xy are chosen at [#permalink]
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30 Apr 2012, 13:34
How come we are assuming that we will choose both expressions at once? That is, why not assume that we choose one of the four expressions, then choose the other one?
For example, 4 letters numbered 1 2 3 4, two of which are chosen at random, what is the probability that the chosen letters will be odd numbered?
Don't these two problems have the same concept?



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Re: If two of the four expressions x+y, x+5y, xy, and 5xy are chosen at [#permalink]
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01 May 2012, 00:32



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Re: If two of the four expressions x+y, x+5y, xy, and 5xy are chosen at [#permalink]
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01 May 2012, 00:51
Bunuel wrote: Lstadt wrote: How come we are assuming that we will choose both expressions at once? That is, why not assume that we choose one of the four expressions, then choose the other one?
For example, 4 letters numbered 1 2 3 4, two of which are chosen at random, what is the probability that the chosen letters will be odd numbered?
Don't these two problems have the same concept? Mathematically the probability of picking two expressions simultaneously (at once), or picking them one at a time (without replacement) is the same.Hello Bunuel, Maybe telling you how I solved the problem (and got a wrong answer) will explain how I was confused. Since the wanted product is (x + y ) (x  y), I assumed that I will be picking (x+y) then (xy) [from the expressions) which means choosing 1/4 then choosing 1/3 and multiplying them to get 1/12 which is wrong. I think I approached it the same was as 144144 (posted above) did. Thank you very much Bunuel.



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Re: If two of the four expressions x+y, x+5y, xy, and 5xy are chosen at [#permalink]
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01 May 2012, 00:57
Lstadt wrote: Bunuel wrote: Lstadt wrote: How come we are assuming that we will choose both expressions at once? That is, why not assume that we choose one of the four expressions, then choose the other one?
For example, 4 letters numbered 1 2 3 4, two of which are chosen at random, what is the probability that the chosen letters will be odd numbered?
Don't these two problems have the same concept? Mathematically the probability of picking two expressions simultaneously (at once), or picking them one at a time (without replacement) is the same.Hello Bunuel, Maybe telling you how I solved the problem (and got a wrong answer) will explain how I was confused. Since the wanted product is (x + y ) (x  y), I assumed that I will be picking (x+y) then (xy) [from the expressions) which means choosing 1/4 then choosing 1/3 and multiplying them to get 1/12 which is wrong. I think I approached it the same was as 144144 (posted above) did. Thank you very much Bunuel. It should be 1/4*1/3 *2, since you can choose (x+y) first and then (xy) or vise versa.
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Re: If two of the four expressions x+y, x+5y, xy, and 5xy are chosen at [#permalink]
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23 Jul 2012, 11:29
This is how I analysed this. xy is a must here since the form asked is x^2by^2. So probabability of selecting xy is 1/4 and probability of selecting the rest is 2/3. the probability of both is 1/4*2/3. Is this right???
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Re: If two of the four expressions x+y, x+5y, xy, and 5xy are chosen at [#permalink]
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Updated on: 24 Jul 2012, 09:43
144144 wrote: I understand perfectly the 4C2 way, but i want to understand the others as well... . I was hoping someone can help me understand this. What is the 4C2 way? C means what? How I thought of the problem was each equation has a 1/4 chance 1/4 1/4 1/4 1/4  Two out of the 4 equations = success so the probability of getting one of the equations we need is 2/4 or 1/2 then we have 3 equations left over, and the probability of getting the second equation we need is 1/3 so the probability of getting both equations is 1/2*1/3 = 1/6 However I only got this after you guys said that we needed a x+y, xy The b in the question threw me off, I am not sure that if I was given a similar problem some weeks from now I would understand what to do with that b
Originally posted by akuma86 on 23 Jul 2012, 12:01.
Last edited by akuma86 on 24 Jul 2012, 09:43, edited 1 time in total.



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Re: If two of the four expressions x+y, x+5y, xy, and 5xy are chosen at [#permalink]
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23 Jul 2012, 13:47
Akuma,
4C2 stands for binomial coefficient with parameters 4 and 2. In layman's terms, it says how many unique combinations of two can I derive from a set of 4 items. In mathematical terms, it simply means 4! / (2! x (42)!) which simplifies to 4! / (2! x 2!) which is equivalent to 6. Now you just need to figure out how many of the combinations fulfill your criteria  in this case 1  and place that number over 6 to get the probability. To get the amount of unique pairings of Y numbers in a set of X (where order does not matter) I believe you can always use this shorthand formula



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Re: If two of the four expressions x+y, x+5y, xy, and 5xy are chosen at [#permalink]
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09 Nov 2012, 22:18
wonderwhy wrote: If 2 of the 4 expressions x+y, x+5y, xy, 5xy are chosen at random, what is the probability that the product will be in the form of x square i (b*y) square if b is an integer?
Answer options: 1/2 1/3 1/4 1/5 1/6
How is this computed? Total No. of ways of selecting 2 items from 4 items = 4C2 = 6. We can see from the 4 items given that only one pair when multiplied will be of the form (x+ay)(xay). So answer is 1/6. E. Kudos Please... If my post helped.
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Re: If two of the four expressions x+y, x+5y, xy, and 5xy are chosen at [#permalink]
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04 Dec 2012, 17:45
VERY SIMPLE WAY TO SOLVE THIS PROB WITHOUT ANY COMPLEX THEORY N ANY CALCULATION : FROM THE QUESTION STEM , YOU COULD UNDERSTAND VERY EASILY THAT ONLY ONE COMBINATION I.E (X+Y) AND (XY) IS THERE TO GET X2(BY)2 SO WE HAVE TO CHOOSE ONLY (X+Y) AND (XY) wE CAN SELECT IT IN TWO WAY I.E . (X+Y) AND THEN (xY) OR (XY) AND THEN (X+Y) FOR FIRST CASE PROBABILITY =(1/4*1/3) FOR SECOND CASE PROBABILITY =(1/4*1/3) sO TOTAL PROBABILITY =2*((1/4*1/3)=1/6 HOPE IT COULD SIMPLY WORK FOR YOU



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Re: If two of the four expressions x+y, x+5y, xy, and 5xy are chosen at [#permalink]
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07 Dec 2012, 04:34
how many ways to select a pair from 4 expressions? \(\frac{4!}{2!2!}=6\) how many pairs to achieve \(x^2(by)^2\)? (x+y)(xy) only or 1 pair \(\frac{desired}{{all pair possibilities}}=\frac{1}{6}\)
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Re: If two of the four expressions x+y, x+5y, xy, and 5xy are chosen at [#permalink]
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15 Jun 2013, 13:16
I made a careless error on this problem...
I thought that b(y^2) would be the same as (by)^2.... but nope... i didn't catch the parentheses.



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Re: If two of the four expressions x+y, x+5y, xy, and 5xy are chosen at [#permalink]
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30 Jun 2013, 14:44
Great explanations in this topic. Again, when explained, easier than it appeared.
As there are 2 steps to this (number of possible outcomes, number of valid pairs); if you're lost at figuring out the latter it is maybe of some use that it is easy to realize that if there are 6 possible outcomes, only 1/2, 1/3 and 1/6 could be valid answers as x/6 can't be simplified to 1/4 or 1/5. Now you only need to figure out if there are 1, 2 or 3 valid pairs, or if time ran out guess 1 out of 3.



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Re: If two of the four expressions x+y, x+5y, xy, and 5xy are chosen at [#permalink]
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25 Jul 2013, 13:43
I also got 1/6 but not really understanding why the equation pair is (x+y)(xy). My thoughts were that we needed to factor into (xby)(x+by) where the factored "b" was the squared root of the original "b", or essentially a placeholder for a perfect square. This led me to think that the only equation pair possible that would work to satisfy the original ask would be (x+5y) and (xy) since we need two equations with differing signs but also needed an constant infront of the "y" term. Can someone please advise? Thanks, Rich




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