−3(x + 3) = −3(x + 1) − 5.
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P = 10% = 0.1
q = 1 - 0.1 = 0.9
P(at least one defective calculator) = P(1) + P(2) + P(3) + P(4) = 1 - P(0)
The brobability of a binomial distribution is given by
where: n = 4
Therefore,
P(at least one defective calculator) = 1 - 0.6561 = 0.3439
q = 1 - 0.1 = 0.9
P(at least one defective calculator) = P(1) + P(2) + P(3) + P(4) = 1 - P(0)
The brobability of a binomial distribution is given by
where: n = 4
Therefore,
P(at least one defective calculator) = 1 - 0.6561 = 0.3439