document.write( "Question 1197325: Five of 10 new buildings in a city violate the building code. A random sample of three buildings
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document.write( "a)What is the probability of none of the buildings violate the building code in a sample of 3 buildings\r
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document.write( "b)What is the probability of at least one of the new buildings violate the building code in a sample of 3 buildings? \n" );
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Algebra.Com's Answer #830556 by math_tutor2020(3817)![]() ![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "Part (a)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "5 buildings violate code, 10-5 = 5 do not\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "5/10 = probability first selection doesn't violate code \n" ); document.write( "4/9 = probability second selection doesn't violate code \n" ); document.write( "3/8 = probability third selection doesn't violate code\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The numerators decrease: 5,4,3 \n" ); document.write( "So do the denominators: 10,9,8 \n" ); document.write( "This assumes that we're sampling without replacement.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Multiply out the fractions to get this result \n" ); document.write( "(5/10)*(4/9)*(3/8) = (5*4*3)/(10*9*8) \n" ); document.write( "(5/10)*(4/9)*(3/8) = (5*4*3)/(2*5*3*3*2*4) \n" ); document.write( "(5/10)*(4/9)*(3/8) = 1/(2*3*2) \n" ); document.write( "(5/10)*(4/9)*(3/8) = 1/12\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The probability of selecting 3 buildings that don't violate code is 1/12.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "----------------------------------------\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Alternate route:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "n = 5 buildings that don't violate code \n" ); document.write( "r = 3 selections \n" ); document.write( "Use the nCr combination formula \n" ); document.write( "n C r = (n!)/(r!(n-r)!) \n" ); document.write( "5 C 3 = (5!)/(3!*(5-3)!) \n" ); document.write( "5 C 3 = (5!)/(3!*2!) \n" ); document.write( "5 C 3 = (5*4*3!)/(3!*2!) \n" ); document.write( "5 C 3 = (5*4)/(2!) \n" ); document.write( "5 C 3 = (5*4)/(2*1) \n" ); document.write( "5 C 3 = (20)/(2) \n" ); document.write( "5 C 3 = 10 \n" ); document.write( "There are 10 ways to pick three out of the five buildings that are to code.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Using that same formula, you should find that 10C3 = 120 which represents the number of ways to pick any three buildings (whether they are to code or not).\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "As you can probably see, we are using nCr instead of nPr because order doesn't matter. A grouping like ABC is the same as CBA.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "We have 10 ways to pick the three buildings that are to code out of 120 ways to pick the three buildings. \n" ); document.write( "10/120 = 1/12\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "----------------------------------------\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Answer: 1/12\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "======================================================================\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Part (b)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "We'll use the answer of the previous part.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The events \"all 3 don't violate code\" and \"at least one violates code\" are complementary events. \n" ); document.write( "One or the other must happen. \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "P(all 3 don't violate code) + P(at least one violates code) = 1 \n" ); document.write( "P(at least one violates code) = 1 - P(all 3 don't violate code) \n" ); document.write( "P(at least one violates code) = 1 - 1/12 \n" ); document.write( "P(at least one violates code) = 12/12 - 1/12 \n" ); document.write( "P(at least one violates code) = (12 - 1)/12 \n" ); document.write( "P(at least one violates code) = 11/12\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Answer: 11/12 \n" ); document.write( " \n" ); document.write( " |