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:
https://gmatclub.com/forum/probability-q ... en#p643043Hope it's clear.