document.write( "Question 1039599: A study shows that 12% of household appliances break during any given year. If a person has 9 household appliances, what is the probability that exactly 3 of them will break next year? Round your answer to 4 decimal places. \n" ); document.write( "
Algebra.Com's Answer #654367 by jim_thompson5910(35256)\"\" \"About 
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This is a binomial distribution problem. We'll use this formula\r
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\n" ); document.write( "\n" ); document.write( "(n C x)*(p^x)*(1-p)^(n-x)\r
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\n" ); document.write( "\n" ); document.write( "where in this case
\n" ); document.write( "n = 9 is our sample size
\n" ); document.write( "p = 0.12 is the probability of selecting a broken appliance
\n" ); document.write( "x = 3 is the desired number of appliances that break
\n" ); document.write( "n C x is notation referring to the combination formula.\r
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\n" ); document.write( "\n" ); document.write( "First let's use the combination formula to compute n C x = 9 C 3.\r
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\n" ); document.write( "\n" ); document.write( "n C x = (n!)/(x!*(n-x)!)
\n" ); document.write( "9 C 3 = (9!)/(3!*(9-3)!)
\n" ); document.write( "9 C 3 = (9!)/(3!*6!)
\n" ); document.write( "9 C 3 = (9*8*7*6*5*4*3*2*1)/(3*2*1*6*5*4*3*2*1)
\n" ); document.write( "9 C 3 = (362880)/(6*720)
\n" ); document.write( "9 C 3 = (362880)/(4320)
\n" ); document.write( "9 C 3 = 84\r
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\n" ); document.write( "\n" ); document.write( "Plug that in, with the other info, to get...\r
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\n" ); document.write( "\n" ); document.write( "(n C x)*(p^x)*(1-p)^(n-x)
\n" ); document.write( "(9 C 3)*(0.12^3)*(1-0.12)^(9-3)
\n" ); document.write( "(84)*(0.12^3)*(0.88)^(9-3)
\n" ); document.write( "(84)*(0.12^3)*(0.88)^(6)
\n" ); document.write( "(84)*(0.001728)*(0.464404086784)
\n" ); document.write( "0.06740918200488\r
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\n" ); document.write( "\n" ); document.write( "Rounding that to 4 decimal places gives 0.0674\r
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\n" ); document.write( "\n" ); document.write( "The final answer is 0.0674
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