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Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]
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06 Oct 2015, 09:37
Gotcha! Thanks! I guess in this type of questions it's easiest to write out the possibilities?



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Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]
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26 Aug 2016, 09:04
I am poor at probability. Kindly explain where I went wrong: If multiplication of numbers is even means at least one of the two numbers is even.
let a*b = even. a is from first set. so probability of this being even: 2/4 b is from second set. so probability of this being even: 1/3 Accompanied by (or) so I added both.
I know this is wrong but am unable to figure out where I went wrong. Kindly suggest a method in these lines. Thanks



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Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]
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26 Aug 2016, 09:04
I am poor at probability. Kindly explain where I went wrong: If multiplication of numbers is even means at least one of the two numbers is even.
let a*b = even. a is from first set. so probability of this being even: 2/4 b is from second set. so probability of this being even: 1/3 Accompanied by (or) so I added both.
I know this is wrong but am unable to figure out where I went wrong. Kindly suggest a method in these lines. Thanks



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Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]
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22 Oct 2016, 11:25
The best way to solve this is to indetify that the product of being even can be obtained by the following cases:
x=E * y=O = E x=O * y=E = E x=E * y=E = E
The equation above might look quite silly but effectivly when I am saying x=E * y=O I mean that x is even and y is odd. As soon as you find this relationship you did the hard part of the question.
Then is all about replacing the prob/ties: x=E * y=O in GMAT terms can be written as x=E AND y=0 means P(x=E) AND P(y=O) = 2/4 * 2/3 = 4/12 = 1/3 etc...
Correct answer D



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Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]
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21 Nov 2017, 12:11
After seeing these replies I understand why 2/3 is the correct answer. But I still can't figure out why my original logic resulted in an incorrect answer. Can someone please clarify?
I reasoned: If any number in either set is even, then xy will be even. So, probability that even number is selected from first set (2/4) + probability that even number is selected from second set (1/3) = 5/6. (2/4 +1/3=5/6)
Why does this result in the wrong probability?



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Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]
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02 Dec 2017, 18:14
Same question as above! A little confused why that doesn't work. chetan2u can you please shed some light on this? Thanks! pablogutierrez24 wrote: After seeing these replies I understand why 2/3 is the correct answer. But I still can't figure out why my original logic resulted in an incorrect answer. Can someone please clarify?
I reasoned: If any number in either set is even, then xy will be even. So, probability that even number is selected from first set (2/4) + probability that even number is selected from second set (1/3) = 5/6. (2/4 +1/3=5/6)
Why does this result in the wrong probability?
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Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]
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02 Dec 2017, 23:13
Hadrienlbb wrote: Same question as above! A little confused why that doesn't work. chetan2u can you please shed some light on this? Thanks! pablogutierrez24 wrote: After seeing these replies I understand why 2/3 is the correct answer. But I still can't figure out why my original logic resulted in an incorrect answer. Can someone please clarify?
I reasoned: If any number in either set is even, then xy will be even. So, probability that even number is selected from first set (2/4) + probability that even number is selected from second set (1/3) = 5/6. (2/4 +1/3=5/6)
Why does this result in the wrong probability? Hi... What you are doing here is " finding probability of picking an even number from atleast one of the set" Say the set B were 5,6,8 so two even Answer as per the method above 2/4+2/3=7/6 >1!!!! Probability>1? But we are looking for xy to be even.. This will include following cases.. 1) even from A and odd number from B... 2/4 * 2/3=1/3 2) even from A and even from B 2/4 *1/3=1/6 3) odd from A and even from B 2/4*1/3=1/6 Total =1/3+1/6+1/6=2/3 Other easier way would be  Subtract prob of odd from 1 Prob of odd .. odd from both = 2/4*2/3=1/3 So Ans=1 1/3=2/3
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Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]
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03 Dec 2017, 10:35
chetan2u wrote: Hadrienlbb wrote: Same question as above! A little confused why that doesn't work. chetan2u can you please shed some light on this? Thanks! pablogutierrez24 wrote: After seeing these replies I understand why 2/3 is the correct answer. But I still can't figure out why my original logic resulted in an incorrect answer. Can someone please clarify?
I reasoned: If any number in either set is even, then xy will be even. So, probability that even number is selected from first set (2/4) + probability that even number is selected from second set (1/3) = 5/6. (2/4 +1/3=5/6)
Why does this result in the wrong probability? Hi... What you are doing here is " finding probability of picking an even number from atleast one of the set" Say the set B were 5,6,8 so two even Answer as per the method above 2/4+2/3=7/6 >1!!!! Probability>1? But we are looking for xy to be even.. This will include following cases.. 1) even from A and odd number from B... 2/4 * 2/3=1/3 2) even from A and even from B 2/4 *1/3=1/6 3) odd from A and even from B 2/4*1/3=1/6 Total =1/3+1/6+1/6=2/3 Other easier way would be  Subtract prob of odd from 1 Prob of odd .. odd from both = 2/4*2/3=1/3 So Ans=1 1/3=2/3 Thanks for your response! I guess I'm still a bit confused. For example, if a question was asking: "A fair sixsided dice is rolled twice. What is the probability of getting at least once?"Then I understand I could calculate the probability of not getting 3 twice and subtract: \(1  (\frac{5}{6}*\frac{5}{6}) = \frac{11}{36}\) Nevertheless, I could still say: 1. If I get 3 on my first roll I don't care what happen on the 2nd roll = \(\frac{1}{6}*\frac{6}{6}\) 2. If I don't get 3 on the 1st roll, I need 3 on the 2nd = \(\frac{5}{6}*\frac{1}{6}\) 3. Add the two results and get \(\frac{11}{36}\) (Similarly, if the question was asking what is the probability that the sum of 2 rolls is 7, I would do \(1*\frac{1}{6}\)) So here I went about calculating: 1. If I get even from the first set I don't care what the second set gets me = \(\frac{2}{4}*1\) 2. If I get an odd from the first set, I need an even from the second set =\(\frac{2}{4}*\frac{1}{3}\) Now I understand this is wrong, and my first extinct would to subtract the probability of getting two odds. I just don't really understand why this is wrong lol. I'm thinking there is something about dependent / independent events I'm not completely getting...
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If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink]
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11 Dec 2017, 16:50
Bunuel wrote: The Official Guide For GMAT® Quantitative Review, 2ND Edition If x is to be chosen at random from the set {1, 2, 3, 4} and y is to be chosen at random from the set {5, 6, 7}, what is the probability that xy will be even? (A) 1/6 (B) 1/3 (C) 1/2 (D) 2/3 (E) 5/6 Hadrienlbb wrote: chetan2u wrote: Hi...
What you are doing here is " finding probability of picking an even number from atleast one of the set" Say the set B were 5,6,8 so two even Answer as per the method above 2/4+2/3=7/6 >1!!!! Probability>1?
But we are looking for xy to be even.. This will include following cases.. 1) even from A and odd number from B... 2/4 * 2/3=1/3 2) even from A and even from B 2/4 *1/3=1/6 3) odd from A and even from B 2/4*1/3=1/6
Total =1/3+1/6+1/6=2/3
Other easier way would be  Subtract prob of odd from 1 Prob of odd .. odd from both = 2/4*2/3=1/3 So Ans=1 1/3=2/3 Thanks for your response! I guess I'm still a bit confused. . . . So here I went about calculating: 1. If I get even from the first set I don't care what the second set gets me = \(\frac{2}{4}*1\)2. If I get an odd from the first set, I need an even from the second set =\(\frac{2}{4}*\frac{1}{3}\) Now I understand this is wrong, and my first extinct would to subtract the probability of getting two odds. I just don't really understand why this is wrong lol. [I don't think it IS wrong.] I'm thinking there is something about dependent / independent events I'm not completely getting... Hadrienlbb , I think the part in red is not accurate. I am not sure where you think your mistake is, because if I finish your math, the answer is correct. Here is the math from your steps, finished: Quote: 1. If I get even from the first set I don't care what the second set gets me = \(\frac{2}{4}*1\) \(= \frac{1}{2}\)
2. If I get an odd from the first set, I need an even from the second set =\(\frac{2}{4}*\frac{1}{3}\) \(= \frac{2}{12} = \frac{1}{6}\) Now add the results of your numbers 1 and 2: \((\frac{1}{2} + \frac{1}{6}) = (\frac{3}{6} + \frac{1}{6})= (\frac{4}{6}) = \frac{2}{3}\) You just compressed chetan2u 's first two steps, which, with the third (and your second), he adds. Here are his first two of three steps: 1) even from A and odd number from B... 2/4 * 2/3=1/3 2) even from A and even from B 2/4 *1/3=1/6 REWRITE: (2/4 * 2/3) + (2/4 * 1/3) = 2/4 * (2/3 + 1/3) = 2/4 * (1)  which is exactly what you calculated. That's the result for picking an even number from set A. Then add the result for picking an odd number from set A (which both you and he computed identically) \((2/4 * 1) + \frac{1}{6} =\) \((\frac{1}{2} + \frac{1}{6}) = \frac{6 + 2}{12}= \frac{8}{12} =\frac{2}{3}\) What you did here is not the same as the person you were quoting . . . (and I had to delete all that because the machine hollers when there are too many quotes within quotes). Maybe I am missing something. Now I am the confused one, I think. It seems to me you got it right. Why do you think there is a mistake?
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If x is to be chosen at random from the set {1, 2, 3, 4} and
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