document.write( "Question 1161643: Every day, there is a 90% chance that Kevin will spill coffee on himself while driving to work.\r
\n" ); document.write( "\n" ); document.write( "Looking at the next eight days, what is the probability that Kevin will spill coffee on himself on just two of those days? \r
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\n" ); document.write( "\n" ); document.write( "Looking at the next eight days, what is the probability that Kevin will spill coffee on himself on at least two of those days?
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Algebra.Com's Answer #785250 by KMST(5347)\"\" \"About 
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Each day Kevin drives to work,
\n" ); document.write( "there is a 90% chance (p=0.9) that Kevin will spill coffee on himself while driving, and
\n" ); document.write( "there is a 100%-90%=10% chance (p=1-0.9=0.1) that Kevin will not spill coffee on himself while driving.
\n" ); document.write( "If you consider one day that Kevin drives to work, the binomial
\n" ); document.write( "\"%280.9s%2B0.1n%29\" represents that p=1 possibility that one or another of the possible outcomes will happen.
\n" ); document.write( "The coefficient \"0.9\" in front of \"s\" (for spill) represents the possibility that Kevin will spill coffee on himself.
\n" ); document.write( "The coefficient \"0.1\" in front of \"n\" (for no spill) represents the possibility that Kevin will not spill coffee on himself.
\n" ); document.write( "considering multiple days, the possibilities multiply.
\n" ); document.write( "If you consider \"8\" days that Kevin drives to work, the power of a binomial
\n" ); document.write( "\"%280.9s%2B0.1n%29%5E8\"
\n" ); document.write( "\"%22=%22\"
\n" ); document.write( "represent the probability that one or another of the many outcomes possible will happen.
\n" ); document.write( "The coefficient of the term
\n" ); document.write( "\"28%280.9%5E2s%5E2%29%280.1%5E6n%5E6%29=28%280.9%5E2%29%280.1%5E6%29s%5E2n%5E6\"
\n" ); document.write( "represents the possibility that
\n" ); document.write( "Kevin will spill coffee on himself on 2 of those days,
\n" ); document.write( "but will not spill coffee on himself on the other 6 days.
\n" ); document.write( "That is
\n" ); document.write( "\"28%280.9%5E2%29%280.1%5E6%29=28%2A0.81%2A0.000001=0.00000081\" or
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\n" ); document.write( "The possibility that Kevin will not spill coffee on himself while driving on any of those 8 days is the coefficient o the term
\n" ); document.write( "\"0.1%5E8n%5E8\" , which is \"0.1%5E8=1%2A10%5E-0.1%2A10%5E-7\" or \"0.00000001\".
\n" ); document.write( "The possibility that Kevin will spill coffee on himself while driving only 1 of those 8 days is the coefficient o the term
\n" ); document.write( "\"8%280.9s%29%280.1%5E7s%5E7%29=8%2A0.9%2A0.1%5E7sn%5E7\" , which is
\n" ); document.write( "\"8%2A0.9%2A0.1%5E7=7.2%2A10-7\" or \"0.00000072\" .
\n" ); document.write( "Adding those two possibilities,
\n" ); document.write( "the possibility that Kevin will sill coffe on himself on no more than 1 of those 8 days is
\n" ); document.write( "\"7.2%2A10%5E-7%2B0.1%2A10%5E-7=7.3%2A10%5E-7\" or \"0.00000072%2B0.00000001=0.00000073\" .
\n" ); document.write( "The possibility of the opposite (Kevin spilling coffee on himself at least 2 of the 8 days is
\n" ); document.write( "\"1-0.00000073=0.99999927\" .
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