1) Suppose that 1.68 percent of the products manufactured at by a company are de
ID: 3044081 • Letter: 1
Question
1) Suppose that 1.68 percent of the products manufactured at by a company are defective. If a product from this company is defective, it will break within the first year with probability 0.901. If a product from this company is not defective, it will break within the first year with probability 0.015. What is the probability that a product from this company will break within the first year?
2)Suppose that 1.08 percent of the products manufactured at by a company are defective. If a product from this company is defective, it will break within the first year with probability 0.948. If a product from this company is not defective, it will break within the first year with probability 0.028. What is the probability that a product from this company was defective, conditional on the event that it breaks within the first year?
3)In a particular digital communication system, the signal transmitted over a bit interval is 0 with probability 0.49 and 1 with probability 1 - 0.49. If the transmitted signal is a 0, the probability that it will be erased is 0.013, and if the transmitted signal is a 1, the probability that it will be erased is 0.075. What is the probability that a transmitted signal will be erased?
4)In a particular digital communication system, the signal transmitted over a bit interval is 0 with probability 0.548 and 1 with probability 1 - 0.548. If the transmitted signal is a 0, the probability that it will be erased is 0.021, and if the transmitted signal is a 1, the probability that it will be erased is 0.073. What is the probability that 0 was transmitted, conditional on the event that the signal was erased?
Round your answer to 3 significant digits.
Explanation / Answer
1)
P(D) =0.0168
P(B|D)= 0.901
P(B|ND)=0.015
P(ND) = 1 -P(D) = 0.9832
P(B)=P(B D) + P(B ND)
= P(D)* P(B|D) + P(ND)*P(B|ND)
= 0.0168* 0.901 + 0.9832* 0.015
= 0.0151368 + 0.014748
= 0.0298848
= 0.030
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