A student needs eight chips of a certain type to build a circuit. It is known that 5% of these chips are defective. How many chips should he buy for there to be a greater than 90% probability of having enough chips for the circuit? (Use Binomial Probability Law and consider n = 8, 9, ...)
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Expert's answer
2019-07-30T10:04:46-0400
Binomial probability law:
"P(k,n)=\\frac{n!}{k!(n-k)!}p^k(1-p)^{n-k}"
p=0.95.
To determine the probability that the working chips are greater than seven, we add the probabilities for each k starting from 8 and ending with n.
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