SOLUTION: 3% of total bulbs are defective from past records.find the probability that the number of defective bulbs are 0 or 1 or 2 or 3 or 4 or 5 if the total number of bulbs are 100
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Question 252673: 3% of total bulbs are defective from past records.find the probability that the number of defective bulbs are 0 or 1 or 2 or 3 or 4 or 5 if the total number of bulbs are 100 Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! 3% of total bulbs are defective from past records.find the probability that the number of defective bulbs are 0 or 1 or 2 or 3 or 4 or 5 if the total number of bulbs are 100
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Use the binomial formla for 100 trials with p = 0.03 and x = 0,
then x = 1, then x = ........and x=5 :
P(x=k) = nCk*p^k*q(n-k)
Ex:
P(x= 2) = 100C2*(0.03)^2*0.97^8 = 0.2252..
Cheers,
Stan H.
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