document.write( "Question 40007: The city water department estimates that for any day in January, the probability of a water main freezing and breaking is 2. For six consecutive days no water mains break. What is the probability that this will happen?\r
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Algebra.Com's Answer #25417 by AnlytcPhil(1806)\"\" \"About 
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The city water department estimates that for any day in January, the\r\n" );
document.write( "probability of a water main freezing and breaking is 2.  For six consecutive\r\n" );
document.write( "days no water mains break.  What is the probability that this will happen?\r\n" );
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document.write( "First of all, no probability of anything could ever be 2.  All probabilities are\r\n" );
document.write( "between 0 and 1, never 2.  Perhaps you meant .2, not 2.\r\n" );
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document.write( "If the probability that a water main breaks is .2, then the probability that\r\n" );
document.write( "none break is 1 - .2 = .8.\r\n" );
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document.write( "Assuming water mains breaking are independent events, then we cal multiply their\r\n" );
document.write( "probabilities.\r\n" );
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document.write( "P(no main breaks the 1st day & no main breaks the 2nd day & no main breaks the\r\n" );
document.write( "3rd day & no main breaks the 4th day & no main breaks the 5th day & no main\r\n" );
document.write( "breaks the 6th day) =  \r\n" );
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document.write( "P(no main breaks the 1st day) × P(no main breaks the 2nd day) × P(no main\r\n" );
document.write( "breaks the 3rd day) × P(no main breaks the 4th day) × P(no main breaks the 5th\r\n" );
document.write( "day) × P(no main breaks the 6th day) = \r\n" );
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document.write( "(.8)(.8)(.8)(.8)(.8)(.8) = (.8)6 = .262144\r\n" );
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document.write( "Edwin

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