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There are 10 balls numbered 1-10 and two balls are picked WITH and WIT

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There are 10 balls numbered 1-10 and two balls are picked WITH and WIT  [#permalink]

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New post 19 Dec 2010, 11:31
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There are 10 balls numbered 1-10 and two balls are picked WITH and WITHOUT replacement (2 cases). What is the probability that their sums are ODD or EVEN?

Without working the numbers I feel it should be 50%. However can someone (Bunuel) work the math and show how it works?

When I do the math, say for the case without replacement:
2 balls out of 10 -> 10C2 = 45
Even sum when the balls are both even or both odd. This can happen as 2 x 5C2 = 20
So probability = 20/45

Where am I wrong?

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Re: There are 10 balls numbered 1-10 and two balls are picked WITH and WIT  [#permalink]

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New post 19 Dec 2010, 11:37
mainhoon wrote:
There are 10 balls numbered 1-10 and two balls are picked WITH and WITHOUT replacement (2 cases). What is the probability that their sums are ODD or EVEN?

Without working the numbers I feel it should be 50%. However can someone (Bunuel) work the math and show how it works?

When I do the math, say for the case without replacement:
2 balls out of 10 -> 10C2 = 45
Even sum when the balls are both even or both odd. This can happen as 2 x 5C2 = 20
So probability = 20/45

Where am I wrong?


You're not wrong!

To get an even sum, you must add two odds or two evens; to get an odd sum, you add an odd and an even.

So, in the "without replacement" case, you have a better chance to get an odd sum.

Let's review the 4 cases:

odd then even: 5/10 * 5/9 = 25/90
even then odd: 5/10 * 5/9 = 25/90

odd then odd: 5/10 * 4/9 = 20/90
even then even: 5/10 * 4/9 = 20/90

As you can see, there's a 50/90 chance of getting an odd sum and only a 40/90 chance of getting an even sum.

In the "with replacement" case the odds are 50/50, since each of the 4 cases breaks down to:

5/10 * 5/10 = 25/100.
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Re: There are 10 balls numbered 1-10 and two balls are picked WITH and WIT  [#permalink]

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New post 19 Dec 2010, 11:57
Ah thanks for the quick reply. I guess I was thinking conceptually why should odd be favored over even. In the with replacement case so it is 50-50 to get an odd or even sum, but in the without replacement case, odds for odd are higher than even.. Well a followup.

I solved the question with the combination formula for the even-without-replacement case. Can you show how to solve this with the combination formula with the odd-without-replacement case?

2 of 10 balls = 10C2 = 45
Odd Sum: Odd-Even or Even-Odd: 5C1 x 5C1 + 5C1 x 5C1 = 50? [This should really be 25]

What is wrong here?
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Re: There are 10 balls numbered 1-10 and two balls are picked WITH and WIT  [#permalink]

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New post 19 Dec 2010, 20:36
mainhoon wrote:
Ah thanks for the quick reply. I guess I was thinking conceptually why should odd be favored over even. In the with replacement case so it is 50-50 to get an odd or even sum, but in the without replacement case, odds for odd are higher than even.. Well a followup.

I solved the question with the combination formula for the even-without-replacement case. Can you show how to solve this with the combination formula with the odd-without-replacement case?

2 of 10 balls = 10C2 = 45
Odd Sum: Odd-Even or Even-Odd: 5C1 x 5C1 + 5C1 x 5C1 = 50? [This should really be 25]

What is wrong here?


Hi,

since it's a combinations problem, order doesn't matter - there are 5 odd balls and we want one, so there's 5C1 ways to choose; similarly, there are 5 even balls and we want to choose one, so there's 5c1 ways to choose. We're choosing one odd AND one even, so we multiply:

5C1 * 5C1 = 5 * 5 = 25

Your mistake was considering odd then even and even then odd as two different possibilities - that's why you got double the actual answer.
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Re: There are 10 balls numbered 1-10 and two balls are picked WITH and WIT  [#permalink]

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New post 20 Dec 2010, 02:26
mainhoon wrote:
There are 10 balls numbered 1-10 and two balls are picked WITH and WITHOUT replacement (2 cases). What is the probability that their sums are ODD or EVEN?

Without working the numbers I feel it should be 50%. However can someone (Bunuel) work the math and show how it works?

When I do the math, say for the case without replacement:
2 balls out of 10 -> 10C2 = 45
Even sum when the balls are both even or both odd. This can happen as 2 x 5C2 = 20
So probability = 20/45

Where am I wrong?


WITHOUT REPLACEMENT CASE



Combinatorial approach:
Odd sum: \(P(eo)+P(oe)=\frac{C^1_5*C^1_5+C^1_5*C^1_5}{C^1_{10}*C^1_{9}}=\frac{50}{90}\) or which is the same \(P(odd \ sum)=2*\frac{C^1_5*C^1_5}{C^1_{10}*C^1_{9}}=\frac{50}{90}\);

Note here that \(C^2_{10}\) won't give you the # of total outcomes (denominator) as C here stands for combinations and by definition it gives the # of selections of 2 different objects out of 10 when order of the selection doesn't matter, but total # of outcomes here is 10*9=90 (1-2, 1-3, 1-4, ..., 2-1, ..., 10-9) so for total # of outcomes order matters as for example 2-4 is different outcome than 4-2.

Here you can solve the question so that to have \(C^2_{10}\) in the denominator. Though in this case \(C^2_{10}\) will represent all possible pairs of 2 numbers out of 10, some will give odd sum and other even. Thus nominator also should be based on the same approach: \(C^1_5*C^1_5\) will give all possible pairs of odd and even numbers (as we are interested in a selection of odd-even pair then the order doesn't matter), so \(P=\frac{C^1_5*C^1_5}{C^2_{10}}=\frac{25}{45}=\frac{50}{90}\).

Even sum: \(P(ee)+P(oo)=\frac{C^1_5*C^1_4+C^1_5*C^1_4}{C^1_{10}*C^1_{9}}=\frac{40}{90}\);

Here you can also solve the question so that to have \(C^2_{10}\) in the denominator. Again \(C^2_{10}\) will represent all possible pairs of 2 numbers out of 10, some will give odd sum and other even. Thus nominator also should be based on the same approach: \(C^2_5\) will give all possible pairs of 2 even numbers out of 5 and \(C^2_5\) will give all possible pairs of 2 odd numbers out of 5 (as both will give even sum), so \(P=\frac{C^2_5+C^2_5}{C^2_{10}}=\frac{20}{45}=\frac{40}{90}\).

Probability approach:
Odd sum: \(P(eo)+P(oe)=\frac{5}{10}*\frac{5}{9}+\frac{5}{10}*\frac{5}{9}=\frac{50}{90}\) or which is the same \(P(odd \ sum)=2*\frac{5}{10}*\frac{5}{9}=\frac{50}{90}\), multiplying by 2 as odd-even can occur in two ways OE and EO;

Even sum: \(P(ee)+P(oo)=\frac{5}{10}*\frac{4}{9}+\frac{5}{10}*\frac{4}{9}=\frac{40}{90}\);


WITH REPLACEMENT CASE


No matter what ball you pick for the first one, you'll have 1/2 chances to make the sum even with the second ball and the same 1/2 chances to make the sum odd.

Similar problem: http://gmatclub.com/forum/probability-q ... en#p643043

Hope it's clear.
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Re: There are 10 balls numbered 1-10 and two balls are picked WITH and WIT  [#permalink]

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New post 21 Dec 2010, 06:24
skovinsky wrote:
mainhoon wrote:
In the "with replacement" case the odds are 50/50, since each of the 4 cases breaks down to:

5/10 * 5/10 = 25/100.


Hi
I don´t understand what you want to say here.
For the "with replacement" scenario I simply have 2 different cases:
when first is odd, second is even \(\frac{5}{10}*\frac{5}{10}\)
when first is even, second is odd \(\frac{5}{10}*\frac{5}{10}\)

That is \(\frac{5}{10}*\frac{5}{10}*2=\frac{1}{2}\)
Is that what you mean?
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Re: There are 10 balls numbered 1-10 and two balls are picked WITH and WIT  [#permalink]

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New post 21 Dec 2010, 18:29
medanova wrote:
skovinsky wrote:
mainhoon wrote:
In the "with replacement" case the odds are 50/50, since each of the 4 cases breaks down to:

5/10 * 5/10 = 25/100.


Hi
I don´t understand what you want to say here.
For the "with replacement" scenario I simply have 2 different cases:
when first is odd, second is even \(\frac{5}{10}*\frac{5}{10}\)
when first is even, second is odd \(\frac{5}{10}*\frac{5}{10}\)

That is \(\frac{5}{10}*\frac{5}{10}*2=\frac{1}{2}\)
Is that what you mean?


Hi,

I outlined 4 cases in my post (o/o, o/e, e/o, e/e). In the "with replacement" scenario, each is equally likely (i.e. has a 5/10 * 5/10 = 25/100 chance to happen). Since 2 of the 4 cases match what we want, there's a 25/100 + 25/100 = 50/100 probability of getting what we desire.
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Re: There are 10 balls numbered 1-10 and two balls are picked WITH and WIT  [#permalink]

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Re: There are 10 balls numbered 1-10 and two balls are picked WITH and WIT &nbs [#permalink] 14 Dec 2017, 04:49
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