hw1_soln_1_2 - OR360 Problem Set 1 Solutions to Problems 1...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: OR360 Problem Set 1 Solutions to Problems 1 and 2 1. Use the definition of the "choose" notation and compute directly, starting from the right-hand side. Bring the two fractions to a common denominator so they can be added, then simplify: n-1 n-1 + r-1 r = = = (n - 1)! (n - 1)! + (r - 1)!(n - r)! r!(n - r - 1)! (n - 1)!r + (n - 1)!(n - r) r!(n - r)! n! n (n - 1)!n = = r!(n - r)! r!(n - r)! r For a combinatorial argument, consider a particular element X in a group of n elements. A subset of r elements out of this group either contains X or it does not, hence the total number of subsets equals the sum of the number of subsets containing X and the number not containing X. In order to choose a subset containing X, we take X into the subset, and choose another r - 1 elements out of the remaining n - 1 elements in the group, so there are n-1 ways to do r-1 this. In order to choose a subset not containing X, we place X aside and choose the r elements out of the remaining n - 1 elements, so there are n-1 ways to do this. r 2. There are many different ways to prove this. The easiest to note is that if A, B, C are independent, then A is independent from any set formed with B, C (this is proved in Ross, p.78). This then implies that Ac is independent from any set formed by B, C by Prop. 3.8.1 (Ross, p.77), and thus in particular, Ac and B C c are independent. This can also be done without using the theorems in the book. Rewrite the left hand side of the equation as: LHS = P {Ac (B C c )} = P {Ac B C c } = P {B} - P {B A} - P {B C} + P {A B C} = P {B} (1 - P {A} - P {C} + P {A} P {C} = P {B} (1 - P {A})(1 - P {C}) Meanwhile, the right hand side can be rewritten as: RHS = P {Ac } P {B C c } = (1 - P {A})(P {B} - P {B C}) = (1 - P {A})(P {B} - P {B} P {C}) = (1 - P {A})P {B} (1 - P {C}) . We see that the two sides are equal. (independence) (inclusion-exclusion) (independence) = P {B} - P {B} P {A} - P {B} P {C} + P {A} P {B} P {C} 1 ...
View Full Document

This note was uploaded on 09/22/2009 for the course ORIE 3500 taught by Professor Weber during the Summer '08 term at Cornell University (Engineering School).

Ask a homework question - tutors are online