document.write( "Question 438077: the probability of making a mistake on any given day is 3/4. If I have an opportunity to make a mistake each day one per day for the next 7 days what is the probability I will make a mistake in at least 2 opportunities??
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document.write( "Help please...test is due and I haven't a clue!
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document.write( "Thank You Thank You!
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document.write( "joeboohoo \n" );
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Algebra.Com's Answer #303028 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! the probability of making a mistake on any given day is 3/4. If I have an opportunity to make a mistake each day one per day for the next 7 days what is the probability I will make a mistake in at least 2 opportunities?? \n" ); document.write( "----- \n" ); document.write( "Its binomial because you either make a mistake or you don't. \n" ); document.write( "--- \n" ); document.write( "n = 7 ; p = 3/4 \n" ); document.write( "---- \n" ); document.write( "Each day you have probability of 3/4 to make a mistake and 1/4 to not \n" ); document.write( "make a mistake. \n" ); document.write( "--- \n" ); document.write( "P(2<= x <=7) = 1 - P(0<= x <=1) = 1 - binomcdf(7,3/4,1) = 0.9987 \n" ); document.write( "================ \n" ); document.write( "cheers, \n" ); document.write( "Stan H. \n" ); document.write( " |