The first automatic machine gives 1% of scrap, the second – 1.5%, and the third – 2%. One part was randomly
The first automatic machine gives 1% of scrap, the second – 1.5%, and the third – 2%. One part was randomly selected from each machine. What is the probability that the following will be standard: a) three details; b) two details; c) at least one detail?
The probability for each of the machines to produce a normal product is:
P1 = 0.99
P2 = 0.985
P3 = 0.98
In order to calculate the probability that each of the selected three parts will be normal, you need to multiply the probabilities P1, P2, P3.
P (A) = P1 * P2 * P3 = 0.99 * 0.985 * 0.98 = 0.955647
The case when two parts turned out to be normal, and one defective can be divided into 3. First – the first part turned out to be defective (P11 = 0.01), and the rest of the parts are normal. The second – the part from the second machine turned out to be defective (P22 = 0.015), the rest are normal. The third – only the part from the third machine turned out to be defective (P33 = 0.02).
P (B) = P11 * P2 * P3 + P1 * P22 * P3 + P1 * P2 * P33 = 0.01 * 0.985 * 0.98 + 0.99 * 0.015 * 0.98 + 0.99 * 0.985 * 0 , 02 = 0.009653 + 0.014553 + 0.019503 = 0.043709
At least one detail should turn out to be normal if all three machines did not result in a defect.
P (C) = 1 – P11 * P22 * P33 = 1 – 0.01 * 0.015 * 0.02 = 0.999997
Answer: probabilities a) 0.955647; b) 0.043709; c) 0.999997.
