{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# HW2 - Mathematics Department UCLA T Richthammer winter 09...

This preview shows pages 1–3. Sign up to view the full content.

Mathematics Department, UCLA T. Richthammer winter 09, sheet 2 Jan 09, 2009 Homework assignments: Math 170A Probability, Sec. 1 009. Let S be a sample space with probability law P . Let E, F be events with P ( E ) = 1 / 2, P ( F c ) = 2 / 3 and P ( E F ) = 2 / 3. Use the properties of a probability law to calculate P ( F ), P ( E F ) and P ( E - F ). Answer: P ( F ) = 1 - P ( F c ) = 1 / 3. P ( E F ) = P ( E ) + P ( F ) - P ( E F ) = 1 / 6. P ( E - F ) = P ( E ) - P ( E F ) = 1 / 3. 010. Let S be a sample space with probability law P . Use the properties of a probability law to show that for arbitrary events E 1 , E 2 we have (a) P ( E 1 E 2 ) = P ( E 1 ) + P ( E c 1 E 2 ) (b) P ( E 1 E 2 ) P ( E 1 ) + P ( E 2 ) (c) P (( E 1 - E 2 ) ( E 2 - E 1 )) = P ( E 1 ) + P ( E 2 ) - 2 P ( E 1 E 2 ) Answer: (a) This follows from the addition rule, as the set on the LHS is the disjoint union of the two sets on the RHS. (b) This follows from (a) as E c 1 E 2 E 2 , and so P ( E c 1 E 2 ) P ( E 2 ). (c) This follows by expressing all sets as disjoint unions of E 1 - E 2 , E 2 - E 1 and E 1 E 2 . (Draw a Venn diagram!) 011. Let S be a sample space with probability law P . Use the properties of a probability law to show that for arbitrary events E 1 , E 2 , E 3 we have (a) P ( E 1 E 2 E 3 ) = P ( E 1 ) + P ( E c 1 E 2 ) + P ( E c 1 E c 2 E 3 ) (Hint: Venn diagram.) (b) P ( E 1 E 2 E 3 ) P ( E 1 ) + P ( E 2 ) + P ( E 3 ) (Hint: Use (a)) (c) Generalize (a) and (b) to a union of n sets E 1 , . . . , E n and prove these formulas. Answer: (a) This follows from the addition rule, as the set on the LHS is the disjoint union of the two sets on the RHS. (b) This follows from (a) as E 2 E c 1 E 2 and E 3 E c 1 E c 2 E 3 . (c) P ( E 1 . . . E n ) = P ( E 1 ) + P ( E c 1 E 2 ) + . . . + P ( E c 1 . . . E c n - 1 E n ) P ( E 1 ) + P ( E 2 ) + . . . + P ( E n ). Proof similarly to (a) and (b). 012. Let S be a sample space with probability law P . Use the properties of a probability law to show that for arbitrary events E 1 , E 2 , E 3 we have (a) P ( E 1 E 2 ) P ( E 1 ) + P ( E 2 ) - 1 (b) P ( E 1 E 2 E 3 ) P ( E 1 ) + P ( E 2 ) + P ( E 3 ) - 2 (Hint: Use 011.(b).) Answer: (a) P ( E 1 E 2 ) = P ( E 1 ) + P ( E 2 ) - P ( E 1 E 2 ) P ( E 1 ) + P ( E 2 ) - 1 as P ( ... ) 1. (b) By 011.(b) P (( E 1 E 2 E 3 ) c ) = P ( E c 1 E c 2 E c 3 ) P ( E c 1 ) + P ( E c 2 ) + P ( E c 3 ). Using P ( E c ) = 1 - P ( E ) the result follows.

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

View Full Document
013. Let S be a sample space with probability law P . Write down the inclusion-exclusion formula explicitely (i.e. without using the summation symbol ) for n = 2 and n = 3. Answer: n = 2: P ( E 1 E 2 ) = P ( E 1 ) + P ( E 2 ) - P ( E 1 E 2 ) n = 3: P ( E 1 E 2 E 3 ) = P ( E 1 ) + P ( E 2 ) + P ( E 3 ) - P ( E 1 E 2 ) - P ( E 1 E 2 ) - P ( E 1 E 3 ) - P ( E 2 E 3 ) + P ( E 1 E 2 E 3 ).
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}