hw3sols09 - Stat 330 1 Solution to Homework 3 Independence...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Stat 330 Solution to Homework 3 1 Independence, Baron p. 40 Events A and B are independent. Show, intuitively, and mathematically, that (a) Their complements are also independent. We know, that P ( A B ) = P ( A ) P ( B ) . We want to find out about the probability of the intersection of the complements. Intuitively, the complements should be independent, since the occurrence of the event or the negative of the events should not matter in the case of independence. Mathematically, we get: P ( A B ) = P ( A ) + P ( B ) - P ( A B ) = = P ( A ) + P ( B ) - (1 - P ( A B )) = = P ( A ) + P ( B ) - (1 - P ( A ) · P ( B )) = = 1 - P ( A ) + P ( B ) - 1 + P ( A ) · P ( B )) = = P ( A ) ( 1 - P ( B )) + P ( B ) = P ( A ) · P ( B ) (b) If they are disjoint, then P ( A ) = 0 or P ( B ) = 0. Since we know, that P ( A B ) = P ( A ) P ( B ) the statement follows directly, if we also assume, that the events are disjoint, i.e. P ( A B ) = 0 (c) If they are exhaustive (i.e. A B = Ω), then P ( A ) = 1 or P ( B ) = 1. From 1 = P ( A B ) we get, 1 = P ( A B ) = 1 - P ( A B ) = 1 - P ( A ) · P ( B ) Therefore P ( A ) · P ( B ) = 0 , which implies that either P ( A ) = 0 or P ( B ) = 0 . Then either P ( A ) = 1 or P ( B ) = 1 . (4 points) 2 Bayes’ Theorem (a) Suppose that a barrel contains many small plastic eggs. Some eggs are painted red and some are painted
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 4

hw3sols09 - Stat 330 1 Solution to Homework 3 Independence...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online