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Chapter 2: Problem 63

An injection-molded part is equally likely to be obtained from any one of the eight cavities on a mold. (a) What is the sample space? (b) What is the probability that a part is from cavity 1 or 2 ? (c) What is the probability that a part is from neither cavity 3 nor 4 ?

Short Answer

Expert verified
(a) \( \{C_1, C_2, C_3, C_4, C_5, C_6, C_7, C_8\} \); (b) \( \frac{1}{4} \); (c) \( \frac{3}{4} \).

Step by step solution

01

Identify the Number of Outcomes

The problem states that an injection-molded part can be obtained from any one of eight cavities. Therefore, each outcome in the sample space represents one cavity. Thus, there are a total of 8 outcomes.
02

Define the Sample Space (Part a)

The sample space is the set of all possible outcomes. In this context, this consists of the cavities numbered 1 through 8. Therefore, the sample space is: \[ S = \{ C_1, C_2, C_3, C_4, C_5, C_6, C_7, C_8 \} \]
03

Calculate Probability for Cavity 1 or 2 (Part b)

To find the probability that a part comes from cavity 1 or 2, we need to count the favorable outcomes and divide by the total number of outcomes. There are 2 favorable outcomes (cavity 1 or 2). Thus, the probability is:\[ P(C_1 \text{ or } C_2) = \frac{2}{8} = \frac{1}{4} \]
04

Calculate Probability for Neither Cavity 3 nor 4 (Part c)

To find the probability that a part comes from neither cavity 3 nor 4, we need to count the favorable outcomes again. The favorable outcomes are from cavities 1, 2, 5, 6, 7, and 8 (excluding 3 and 4). That gives us 6 favorable outcomes. Thus, the probability is:\[ P(\text{not } C_3 \text{ and not } C_4) = \frac{6}{8} = \frac{3}{4} \]

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Key Concepts

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

Understanding Sample Space
A sample space is a fundamental concept in probability that represents the set of all possible outcomes in a given scenario.
In the problem of injection-molded parts, each cavity from which a part can be produced is considered a distinct outcome.
This is because each cavity, identified from 1 to 8, has an equal chance of creating a part when we talk about the sample space.

The sample space, denoted by \( S \), is simply the listing of all these possibilities. For our example, it is expressed as:
  • \( S = \{ C_1, C_2, C_3, C_4, C_5, C_6, C_7, C_8 \} \)
This list includes all the potential cavities a part can come from.
Understanding this complete set of outcomes lays the groundwork for calculating various probabilities related to these outcomes.
Identifying Favorable Outcomes
Favorable outcomes refer to the specific outcomes within the sample space that satisfy a given condition in a probability problem.
In this scenario, we look for outcomes that match particular criteria 鈥� either being from specific cavities or not being from certain ones.

For part (b) of the exercise, the favorable outcomes are the parts coming from cavity 1 or 2. Thus, we count:
  • \( ext{Favorable outcomes} = \{ C_1, C_2 \} \)
This results in 2 favorable outcomes.

For part (c), where we need parts not from cavities 3 or 4, we consider:
  • \( ext{Favorable outcomes} = \{ C_1, C_2, C_5, C_6, C_7, C_8 \} \)
Giving us 6 outcomes that meet the required condition.
Recognizing these outcomes is essential for moving on to the probability calculation stage.
Probability Calculation Made Simple
Calculating probability involves determining how likely it is for a favorable outcome to occur out of all possible outcomes in the sample space.
This is typically done using the formula:
  • \( P( ext{Event}) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}} \)

Applying this to our example:
  • For part (b), the probability that a part comes from cavity 1 or 2:
    \[ P(C_1 \text{ or } C_2) = \frac{2}{8} = \frac{1}{4} \]
  • For part (c), the probability that a part comes from neither cavity 3 nor 4:
    \[ P(\text{not } C_3 \text{ and not } C_4) = \frac{6}{8} = \frac{3}{4} \]
The goal is to count the relevant favorable outcomes and divide by the total number of outcomes, resulting in the desired probabilities.
This method provides a clear pathway to understanding likelihoods in various scenarios based on a well-defined sample space.

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Most popular questions from this chapter

A lot of 100 semiconductor chips contains 20 that are defective. (a) Two are selected, at random, without replacement, from the lot. Determine the probability that the second chip selected is defective. (b) Three are selected, at random, without replacement, from the lot. Determine the probability that all are defective.

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