The probability that the bulb can operate 4000 hours is 0.3. What is the probability that exactly one of eight bulbs can operate 4000 hours?

Correct result:

p2 =  19.8 %


p=0.3  C1(8)=(81)=8!1!(81)!=81=8  p2=100 (81) p (1p)81=100 8 0.3 (10.3)81=19.8%

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