In proof testing of circuit boards, the probability that any particular diode wi
ID: 3024299 • Letter: I
Question
In proof testing of circuit boards, the probability that any particular diode will fail is 0.01. Suppose a circuit board contains 200 diodes.
(b) What is the (approximate) probability that at least six diodes will fail on a randomly selected board? (Round your answer to three decimal places.)
(c) If five boards are shipped to a particular customer, how likely is it that at least four of them will work properly? (A board works properly only if all its diodes work. Round your answer to four decimal places.)
Explanation / Answer
B)
Note that P(at least x) = 1 - P(at most x - 1).
Using a cumulative poisson distribution table or technology, matching
u = the mean number of successes = 2
x = our critical value of successes = 6
Then the cumulative probability of P(at most x - 1) from a table/technology is
P(at most 5 ) = 0.983436392
Thus, the probability of at least 6 successes is
P(at least 6 ) = 0.016563608 [ANSWER]
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c)
Note that the probability of x successes out of n trials is
P(x) = u^x e^(-u) / x!
where
u = the mean number of successes = 2
x = the number of successes = 0
Thus, the probability is
P(Work) = P ( 0 ) = 0.135335283
Note that P(at least x) = 1 - P(at most x - 1).
Using a cumulative binomial distribution table or technology, matching
n = number of trials = 5
p = the probability of a success = 0.135335283
x = our critical value of successes = 4
Then the cumulative probability of P(at most x - 1) from a table/technology is
P(at most 3 ) = 0.998504287
Thus, the probability of at least 4 successes is
P(at least 4 ) = 0.001495713 [ANSWER]