Asst3-solved

Asst3-solved - Statistics 427: Homework #3 Solutions to...

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

View Full Document Right Arrow Icon
Statistics 427: Homework #3 Solutions to Assigned Problems 2.76 [6 pts] The probability that an error is made on any given question is 0 . 1, which means that the probability that no error is made is 0 . 9. Let E i = { error made on question i } . Then P ( E i ) = 0 . 1 and P ( E 0 i ) = 0 . 9. We are interested in the probability that no errors are made in ten questions, or P ( E 0 1 ∩···∩ E 0 10 ). Because the E i are assumed to be independent, the E 0 i are also independent, and so P ( E 0 1 ∩ ··· ∩ E 0 10 ) = 10 Y i =1 P ( E 0 i ) = (0 . 9) 10 = 0 . 3487 . The probability that at least one error is made is that same as 1 - P (no errors are made) = 1 - 0 . 9 10 = 0 . 6513. If we abstract this to probability p of an error and n total questions, then the probabilities are (1 - p ) n and 1 - (1 - p ) n . 2.80 [6 pts] Let E i = { component i works } . Then P ( E i ) = 0 . 9 and we are assuming that all components work independently. In order for the system to work, either the “top” system (1/2) or the “bottom system” (3/4) must work. For the top system to work we must have either component 1 or 2 working, i.e. E 1 E 2 . For the bottom system to work, we must have both 3 and 4 working,
Background image of page 1

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

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

This note was uploaded on 04/07/2010 for the course STAT 427 taught by Professor Staff during the Spring '08 term at Ohio State.

Page1 / 4

Asst3-solved - Statistics 427: Homework #3 Solutions to...

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