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Chapter 2: Problem 108

A batch of 500 containers for frozen orange juice contains 5 that are defective. Two are selected, at random, without replacement from the batch. (a) What is the probability that the second one selected is defective given that the first one was defective? (b) What is the probability that both are defective? (c) What is the probability that both are acceptable? Three containers are selected, at random, without replacement, from the batch. (d) What is the probability that the third one selected is defective given that the first and second ones selected were defective? (e) What is the probability that the third one selected is defective given that the first one selected was defective and the second one selected was okay? (f) What is the probability that all three are defective?

Short Answer

Expert verified
(a) \( \frac{4}{499} \); (b) \( \frac{1}{12475} \); (c) \( 0.980 \); (d) \( \frac{3}{498} \); (e) \( \frac{4}{498} \); (f) \( 0.00000012 \).

Step by step solution

01

Calculate Probability of Second Defective, Given First Defective (Part a)

If the first selected container is defective, only 4 defective containers remain out of 499 total. Therefore, the probability that the second selected container is defective is \( \frac{4}{499} \).
02

Calculate Probability of Both Defective (Part b)

First, calculate the probability the first one is defective: \( \frac{5}{500} \). Then, if the first is defective, calculate the probability the second is also defective: \( \frac{4}{499} \). Multiply these probabilities to find the probability both are defective: \( \frac{5}{500} \times \frac{4}{499} = \frac{1}{12475} \).
03

Calculate Probability of Both Acceptable (Part c)

The probability of selecting an acceptable first container is \( \frac{495}{500} \). After this, the probability of selecting an acceptable second container is \( \frac{494}{499} \). The combined probability that both are acceptable is \( \frac{495}{500} \times \frac{494}{499} = \frac{122265}{124750} \approx 0.980 \).
04

Calculate Probability Third Defective, Given First and Second Defective (Part d)

If the first and second containers are defective, three defective containers remain out of 498 total. Therefore, the probability the third container is defective is \( \frac{3}{498} \).
05

Calculate Probability Third Defective, Given First Defective, Second Okay (Part e)

After selecting one defective and one acceptable container, 4 defective containers and 494 acceptable containers remain out of 498 total. The probability the third container is defective is \( \frac{4}{498} \).
06

Calculate Probability All Three Defective (Part f)

Calculate the probability the first one is defective: \( \frac{5}{500} \). Then, if the first is defective, calculate for the second: \( \frac{4}{499} \). For the third one to be defective, the probability is \( \frac{3}{498} \). Multiply these probabilities: \( \frac{5}{500} \times \frac{4}{499} \times \frac{3}{498} = \frac{1}{8291666.67} \approx 0.00000012 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Theory
Probability theory is the mathematical study that measures the likelihood of events occurring. In this exercise, we explore conditional probability, where the chance of an event is considered after certain preconditions are met. For example, in part (a), we ask: "What is the probability that the second container is defective given the first one was defective?" This is a classic case of conditional probability.
To calculate this, we focus on just the remaining containers after one defective has been removed. Initially, there are 5 defective containers out of 500. If the first selected is defective, only 4 defective containers remain. The overall probability then changes since only 499 containers are left.
This exercise also challenges us to find joint probabilities, like in part (b), where both containers need to be defective. For these situations, multiply the individual probabilities: the probability of the first container being defective and then the second being defective, adjusting each time the chosen sample is removed from the batch.
Combinatorics
Combinatorics is the branch of mathematics that deals with counting, arranging, and combinations, all crucial in probability calculations. Here, each selection of containers is without replacement, an important aspect that changes how we apply combinatorial methods.
Selecting without replacement means our sample space alters with each successive pick; the probability outcome for each step is affected by the previous outcomes. For example, in part (c), once a container is removed from our sample set, it cannot be picked again. This directly influences the probability calculations at each subsequent step.
Parts (d), (e), and (f) also delve deeper into how combinatorial calculations come into play when determining the chances of defective containers appearing multiple times consecutively.
Sample Space Analysis
Sample space analysis is central to solving probability problems as it visually or theoretically outlines all possible outcomes of an experiment or trial. In this exercise, the sample space consists of 500 containers initially, impacting how probabilities are calculated for each scenario.
By narrowing down our initial wide array of potential results, our sample space for each step becomes progressively smaller. For example, when one defective is picked, just 4 remain among the residual 499 containers. This subsequently modifies the event space for subsequent picks.
Evaluating sample space with and without favorable outcomes allows us to build a foundation for other probability findings, such as parts (d) and (e), where previous draw outcomes condition our current sample space. Analyzing how outcomes shift by earlier selections showcases the significance of sample space in conditional probabilities.

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Most popular questions from this chapter

A lot of 100 semiconductor chips contains 20 that are defective. Two are selected randomly, without replacement, from the lot (a) What is the probability that the first one selected is defective? (b) What is the probability that the second one selected is defective given that the first one was defective? (c) What is the probability that both are defective? (d) How does the answer to part (b) change if chips selected were replaced prior to the next selection?

A computer system uses passwords that contain exactly eight characters, and each character is one of the 26 lowercase letters \((a-z)\) or 26 uppercase letters \((A-Z)\) or 10 integers \((0-9)\). Let \(\Omega\) denote the set of all possible password, and let \(A\) and \(B\) denote the events that consist of passwords with only letters or only integers, respectively. Suppose that all passwords in \(\Omega\) are equally likely. Determine the following robabilities: (a) \(P\left(A \mid B^{\prime}\right)\) (b) \(P\left(A^{\prime} \cap B\right)\) (c) \(P\) (password contains exactly 2 integers given that it contains at least 1 integer)

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In the design of an electromechanical product, 12 components are to be stacked into a cylindrical casing in a manner that minimizes the impact of shocks. One end of the casing is designated as the bottom and the other end is the top. (a) If all components are different, how many different designs are possible? (b) If seven components are identical to one another, but the others are different, how many different designs are possible? (c) If three components are of one type and identical to one another, and four components are of another type and identical to one another, but the others are different, how many different designs are possible?

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