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ASK THE PROFESSOR FORUM

Course Documents

Chapter 1 - Intro

Chapter 2 - Methods for Describing Sets of Data

Chapter 3 - Probability

Chapter 4 - Discrete Random Variables

Chapter 5 - Normal Random Variables

Chapter 6 - Sampling Distributions

Chapter 7 - Confidence Intervals

Chapter 8 - Tests of Hypothesis: One Sample

Chapter 9 - Confidence Intervals and Hypothesis Tests: Two Samples

Sample Exam I: Chapters 1 & 2

Sample Exam II: Chapters 3 & 4

Sample Exam III: Chapters 5 & 6

Sample Exam IV: Chapters 7 & 8

Ask the Professor Forum

I am having trouble solving this problem: A sample of 4 different calculators is randomly selected from a group containing 13 that are defective and 27 that have no defects. What is the probability that at least one of the calculator is defective.

Posted to STATS 1 on Monday, August 19, 2013   Replies: 1


Professor Mcguckian
08/19/2013
5:05 PM EST

Hi Gail,
Here is how to solve this one: P(at least one is defective) = 1 - P( none of the four are defective) = 1 - P(1st is not defective)*P(second is not defective|first is not defective)* ... =

1 - 27/40*26/39*25/38*24/37 = 0.808

In this problem, some students assume the calculators would be replaced, but that is not justified in the problem.  Unless with replacement is stated, generally you assume without replacement. The phrase four different calculators points to the fact that you do not want to choose the same one more than once.

Professor McGuckian

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