Quote:
Originally Posted by Unknown008 View Post
Ok, let

A denote the event that the first alternator fails andb
B denote the event that the second alternator fails.

(a) The probability that both fail is represented by .

(b) The probability that neither fails (both don't fail) is represented by

where and represent the event where the alternator does not fail and is equal to 1 - P(A) and 1 - P(B) respectively.

(c) The probability that one or the other fails is represented by .

Whenever you have the word 'both', 'neither', 'and', you denote it by intersection and you multiply the sets (here, a set is the probability of an event)

Whenever you have the word 'or', you denote it by union, and you add the sets.

e.g..



so if the events are independent and the probability a given alternator fails is .02 then P(A) = .02, P(B) = .02?

P(AintersectB) = P(A)xP(B) = (.02)(.02) = 0.0004

P(AintersectB w/bar)= P(AintersectB) = 1-0.0004 = 0.9996

P(AunionB) = P(A) + P(B) = .02 + .02 = 0.04

The probability that both alternators fail is .04%
The probability that neither alternator fails is 99.9%
The probability that one or the other fails is 4%

How do I reply to a post?

Please don't do a new thread for the same problem. This is being addressed over here:
https://www.askmehelpdesk.com/math-sciences/probability-intersection-union-457121.html

Please don't do a new thread for the same problem. This is being addressed over here:
https://www.askmehelpdesk.com/math-sciences/probability-intersection-union-457121.html

Sorry, was not sure how to reply to the original post.