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#p643043Hope it's clear.

_________________

New to the Math Forum?

Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:

GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:

PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.

What are GMAT Club Tests?

Extra-hard Quant Tests with Brilliant Analytics