document.write( "Question 752595: Hi there.\r
\n" ); document.write( "\n" ); document.write( "I'm running 9 trials with 5 possible results -- A,B,C,D, and E -- on each trial.\r
\n" ); document.write( "\n" ); document.write( "How would I calculate the odds of Either A or B showing up at least 6 times?\r
\n" ); document.write( "\n" ); document.write( "I tried .4^6 * 1^3 = ~ .4% which seems too low.\r
\n" ); document.write( "\n" ); document.write( "Total number of combinations is 5^9 -- 1953125. Total combinations with at least 6 A or B seems to be 2^6 * 5^3 -- 8000. Again leads to .4% Is it really that low or am I missing something? \n" ); document.write( "
Algebra.Com's Answer #457921 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
I'm running 9 trials with 5 possible results -- A,B,C,D, and E -- on each trial.
\n" ); document.write( "How would I calculate the odds of Either A or B showing up at least 6 times?
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\n" ); document.write( "Binomial Problem with n = 9 and p(A or B) = 2/5
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\n" ); document.write( "P(6<= x <=9) = 1 - P(0<= x <=5) = 1 - binomcdf(9,2/5,5) = 0.0994
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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