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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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08 Dec 2015, 21:14

Hi Darthestvader, To avoid double counting it is better to just count one way. For instance, 1) You choose 2 0r 4 from set1 and any number from set2: probability = 1/2*1 2) You choose 1 or 3 from set1 and 6 from set2: Probability = 1/2*1/3= 1/6 Total Probability= 1/2+1/6= 8/12 = 2/3

DarthestVader wrote:

Hey all, I'm having trouble understanding why my approach doesn't work and I'd really appreciate your help! I picked E because for xy to be even, you need to pick an even number in first set (and you don't care about what you pick in second set), or an even number in second set (and you don't care what you pick in first set). So then the probability of even in first set is 1/2 and second set is 1/3. Why can't I just add the probabilities together to have 5/6? What have I included in the probability to make it 1/6 bigger than the right answer?

_________________

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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, 08:04

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

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, 08:04

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

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, 10:25

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...

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, 11:11

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)

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)

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
_________________

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 six-sided 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...
_________________

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)

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?