A sample of 20 items is selected for inspection. Let □ be the number of defective items in 20items. Suppose the probability of defectiveness for each item is 0.05.XOHN7911P(x)0.35850.37740.18870.05960.01330.00220.00030.00000.00000.0000--Use the above probability table to find the probability that more than 3 of them are defective.0.01330.0010 +0.0001 = 0.00110.0596 + 0.0133 +0.0022 +0.0003 = 0.07540.0133 + 0.0022 +0.0003 = 0.01580.0133 or 0.0022 or 0.0003

The probability distribution table for number of defective items were given.

Answered: A sample of 20 items is selected for…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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