Comments on: Business Statistic? http://youthinside.org/2008/01/02/business-statistic/ All About Shopping Mon, 21 May 2012 22:21:05 +0000 http://wordpress.org/?v=2.3.1 By: Nathan http://youthinside.org/2008/01/02/business-statistic/#comment-1458 Nathan Fri, 04 Jan 2008 05:30:22 +0000 http://youthinside.org/2008/01/02/business-statistic/#comment-1458 Perhaps I've misunderstood the terms in your question and set up my probability table incorrectly, but I think the answer to your question is that the probability that a new car selected at random needs a warranty repair *or* was not manufatured by a U.S-based company is 0.425. -- (Marginal Probability) Probabilty that a new car needs a warranty repair = 0.04 -- (Marginal probability) Probability that a new car was not manufactured in US = 0.40 -- (Joint Probability) Probability that a new car needs warranty repair and was not manufactured in the US = 0.015 To find this solution, I added the marginal probabilities and subtracted the joint probability to find the mutually exclusive conditions (i.e. car needs repair *or* is not manufactured in the US, but not both conditions together). 0.04 + 0.40 - 0.015 = 0.425 Perhaps I’ve misunderstood the terms in your question and set up my probability table incorrectly, but I think the answer to your question is that the probability that a new car selected at random needs a warranty repair *or* was not manufatured by a U.S-based company is 0.425.

– (Marginal Probability) Probabilty that a new car needs a warranty repair = 0.04
– (Marginal probability) Probability that a new car was not manufactured in US = 0.40
– (Joint Probability) Probability that a new car needs warranty repair and was not manufactured in the US = 0.015

To find this solution, I added the marginal probabilities and subtracted the joint probability to find the mutually exclusive conditions (i.e. car needs repair *or* is not manufactured in the US, but not both conditions together).

0.04 + 0.40 - 0.015 = 0.425

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