352IncExcsoln - x 1 x 2 x 3 = 7 x i ≥ ∀ i So P A 1 = ±...

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

View Full Document Right Arrow Icon
MATH 352 Solution to Practice Inclusion / Exclusion Problem If 11 identical items are randomly distributed amongst 3 labelled boxes, what is the probability that box 1 gets no more than 3, box 2 gets no more than 4, and box 3 gets no more than 6 items? Solution: This desired probability is the probability that a nonnegative integer solution to x 1 + x 2 + x 3 = 11 satisfies x 1 3 , x 2 4 , x 3 6. Let A 1 be the condition that x 1 4, A 2 be the condition that x 2 5, and A 3 be the condition that x 3 7. Then the desired probabilitiy is the probability that none of these events occur, ie. 1 - P ( A 1 A 2 A 3 ). To find P ( A 1 ), we need to count the solutions to
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: x 1 + x 2 + x 3 = 7 , x i ≥ ∀ i . So P ( A 1 ) = ± 7+3-1 7 ² / ± 11+3-1 11 ² . Similarly, P ( A 2 ) = ± 6+3-1 6 ² / ± 11+3-1 11 ² , and P ( A 3 ) = ± 4+3-1 4 ² / ± 11+3-1 11 ² . To find P ( A 1 A 2 ), we need to count the solutions to x 1 + x 2 + x 3 = 2 , x i ≥ ∀ i . So P ( A 1 A 2 ) = ± 2+3-1 2 ² / ± 11+3-1 11 ² . Similarly, P ( A 1 A 3 ) = 1, P ( A 2 A 3 ) = 0, and P ( A 1 A 2 A 3 ) = 0. Therefore the desired probability is [ ± 13 11 ²-± 9 7 ²-± 8 6 ²-± 6 4 ² + ± 4 2 ² + 1] / ± 13 11 ² ....
View Full Document

This note was uploaded on 01/15/2010 for the course MATH 352 taught by Professor M.tao during the Spring '09 term at University of Victoria.

Ask a homework question - tutors are online