Question 752595: Hi there.
I'm running 9 trials with 5 possible results -- A,B,C,D, and E -- on each trial.
How would I calculate the odds of Either A or B showing up at least 6 times?
I tried .4^6 * 1^3 = ~ .4% which seems too low.
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?
Answer by stanbon(75887)   (Show Source):
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.
How would I calculate the odds of Either A or B showing up at least 6 times?
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Binomial Problem with n = 9 and p(A or B) = 2/5
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P(6<= x <=9) = 1 - P(0<= x <=5) = 1 - binomcdf(9,2/5,5) = 0.0994
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Cheers,
Stan H.
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