Probability Problem

[GHong14]

Wondering how this problem would differ if there was NO REPLACEMENT

[anshumishra]

Before solving this question I would guess the answer should be 1/2 just because of there should be no reason why odd should have a better/worse probability when the sample space we are having contains (50even,50 odd numbers).

For the current question , there are these ways in which the sum will be odd for the 3 draws :

odd,even,even
odd,odd,odd
even,odd,even
even,even,odd

So, the required probability = 1/2*1/2*1/2* 4 = 1/2

Similarly, I would guess, the probability with replacement of balls, should be 1/2. (why should it matter whether one draws odd ball or even ball in 1st, 2nd or 3rd attempts).

Thanks

[Night reader]

Wondering how this problem would differ if there was NO REPLACEMENT

as the question was not answered, I decided to post

probability with replacement means that every draw made is followed with a replacement (return) of the drawn object into the sample (or population). e.g. there are 4 balls- 1 blue and 3 yellow balls, what is P(yellow ball)?
a) with replacement => the first draw -yellow 3/4, the second draw, the ball is yellow 3/4, the third draw, the ball is yellow 3/4
b) without replacement, what is P(yellow ball)? the first draw -yellow 3/4, the second draw, the ball is yellow 2/3, the third draw, the ball is yellow 1/2

Likewise the example above, given the condition was changed - we have to make selection of three balls with odd and even numbers for the sum to be odd without replacement.

ODD EVEN EVEN => 50/100 * 50/99 * 49/98
ODD ODD ODD => 50/100 * 49/99 * 48/98
EVEN ODD EVEN => 50/100 * 50/99 * 49/98
EVEN EVEN ODD => 50/100 * 49/99 * 50/98

Our probability would result in 1/2; the difference between draws with replacement and without replacement is that such an independence of each draw from another => depending on which draw results in ODD or Even we calculate the new probability for the balls left in our box (sample)

That our probabilities with and without replacement are the same i.e. 1/2 is based on the fact rest on the fundamental probability concept => P=number of favorable outcomes/number of total outcomes
above the number of favorable outcomes for drawing ODD and drawing EVEN are the same 5 even and 5 odd => hence their probabilities with and without replacement would result in the similar probability despite dependence (not independence) of each draw from another.

[anshumishra]

Wondering how this problem would differ if there was NO REPLACEMENT

as the question was not answered, I decided to post

probability with replacement means that every draw made is followed with a replacement (return) of the drawn object into the sample (or population). e.g. there are 4 balls- 1 blue and 3 yellow balls, what is P(yellow ball)?
a) with replacement => the first draw -yellow 3/4, the second draw, the ball is yellow 3/4, the third draw, the ball is yellow 3/4
b) without replacement, what is P(yellow ball)? the first draw -yellow 3/4, the second draw, the ball is yellow 2/3, the third draw, the ball is yellow 1/2

Likewise the example above, given the condition was changed - we have to make selection of three balls with odd and even numbers for the sum to be odd without replacement.

ODD EVEN EVEN => 50/100 * 50/99 * 49/98
ODD ODD ODD => 50/100 * 49/99 * 48/98
EVEN ODD EVEN => 50/100 * 50/99 * 49/98
EVEN EVEN ODD => 50/100 * 49/99 * 50/98

Our probability would result in 1/2; the difference between draws with replacement and without replacement is that such an independence of each draw from another => depending on which draw results in ODD or Even we calculate the new probability for the balls left in our box (sample)

That our probabilities with and without replacement are the same i.e. 1/2 is based on the fact rest on the fundamental probability concept => P=number of favorable outcomes/number of total outcomes
above the number of favorable outcomes for drawing ODD and drawing EVEN are the same 5 even and 5 odd => hence their probabilities with and without replacement would result in the similar probability despite dependence (not independence) of each draw from another.

It was answered in my last para, however I agree that I didn't confirm the answer as I did for the first case(without replacement). Thanks for making it clear. Now I guess, I should have probably posted that. It does help to understand.

Takeaways, which I follow and hence didn't require to solve : Once you find a symmetry, you really don't need to solve for all the different cases. Here the sample space (50 even, 50 odd) is uniquely distributed, now you can apply whatever rules to both of them(for e.g. some weird rule - say- alternate replacement) , they will share equal probabilities.