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A slitter assembly contains 48 blades. Five blades are selected at random and evaluated each day for sharpness. If any dull blade is found, the assembly is replaced with a newly sharpened set of blades. a. If 10 of the blades in an assembly are dull, what is the probability that the assembly is replaced the first day it is evaluated? b. If 10 of the blades in an assembly are dull, what is the probability that the assembly is not replaced until the third day of evaluation? [Hint: Assume that the daily decisions are independent, and use the geometric distribution. \(]\) c. Suppose that on the first day of evaluation, 2 of the blades are dull; on the second day of evaluation, 6 are dull; and on the third day of evaluation, 10 are dull. What is the probability that the assembly is not replaced until the third day of evaluation? [Hint: Assume that the daily decisions are independent. However, the probability of replacement changes every day.

Step by step solution

01

Find Probability of Selection of No Dull Blades

When selecting 5 blades from 48, the probability of choosing only sharp blades is found using combinations. There are 10 dull blades, so 38 are sharp.The probability of selecting only sharp blades is the combination of selecting 5 sharp blades from 38, divided by the combination of selecting any 5 blades from 48:\[ P(\text{sharp}) = \frac{\binom{38}{5}}{\binom{48}{5}} \]

02

Calculate Probability Assembly Is Replaced on First Day

The probability of having at least one dull blade among the 5 selected is found by subtracting the probability from Step 1 from 1:\[ P(\text{dull on day 1}) = 1 - P(\text{sharp}) = 1 - \frac{\binom{38}{5}}{\binom{48}{5}} \]

03

Use Geometric Distribution for Replacement on Third Day

Given that daily decisions are independent, use the geometric distribution to find the probability of the assembly not being replaced until the third day. Thus, it is necessary that the assembly is not replaced on the first two days and gets replaced on the third day:\[ P(\text{not replaced until day 3}) = (1 - P(\text{dull on day 1}))^2 \times P(\text{dull on day 1}) \]

04

Calculate Probabilities for Each Day with Varying Dull Blades

For this part, calculate the probability for the first day (2 dull blades out of 48), second day (6 dull blades), and third day (10 dull blades), assuming independent events. Adjust the probability of choosing sharp blades for each scenario:- Day 1: \( P_1(\text{dull}) = 1 - \frac{\binom{46}{5}}{\binom{48}{5}} \)- Day 2: \( P_2(\text{dull}) = 1 - \frac{\binom{42}{5}}{\binom{48}{5}} \)- Day 3: \( P_3(\text{dull}) = 1 - \frac{\binom{38}{5}}{\binom{48}{5}} \)The probability that the assembly is not replaced until the third day is:\[ (1 - P_1(\text{dull})) \times (1 - P_2(\text{dull})) \times P_3(\text{dull}) \]

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Key Concepts

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

Geometric Distribution

The geometric distribution is a fascinating concept in probability theory. It often deals with scenarios where we're trying to find the probability of achieving the first success after a series of trials. In this exercise, the success event is finding at least one dull blade when evaluating 5 blades out of a total of 48.
By definition, the geometric distribution expresses the likelihood of geometric randomness where each trial is independent and has the same probability of success. When dealing with the situation described, using this makes perfect sense.

• **Independent Trials**: Each day, the decision to replace the assembly doesn't affect future evaluations. This independence allows us to utilize the geometric distribution's properties.
• **Parameter p**: Calculate this by finding the probability of success (finding a dull blade on the first day) using complements, as shown in the solution \[ P(\text{dull on day 1}) = 1 - P(\text{sharp}) = 1 - \frac{\binom{38}{5}}{\binom{48}{5}} \] where \( P(\text{sharp}) \) is the probability of selecting sharp blades.
• **Geometric Probability**: To determine the probability of not replacing the assembly until the third day, we calculate \[ (1 - P(\text{dull on day 1}))^2 \times P(\text{dull on day 1}) \] This captures the requirement of failures (no dull blades on days 1 and 2) followed by a success (finding dull blades on day 3).

Combinatorics

Combinatorics is a branch of mathematics that deals with counting, arranging, and finding patterns. In probability, understanding how to calculate combinations is essential, especially when determining the chances of selecting certain items from a group.
In this exercise, combinatorics helps determine the probability of selecting different numbers of dull blades when sampling 5 blades from the assembly.

• **Combinations Formula**: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \] where \( n \) is the total number of items to choose from and \( k \) is the number of items to select. Factorials (\( n! \)) are used to denote the product of all positive integers up to \( n \).
• **Applications**: In this case, we use combinations to calculate the number of ways \( \binom{38}{5} \) lets us choose 5 sharp blades from 38 sharp ones, while \( \binom{48}{5} \) calculates the total ways to choose any 5 blades from 48.
• **Importance**: It allows determination of the probability of various outcomes using tools like the binomial distribution complement (a fundamental in probability theory) \[ P(\text{sharp}) = \frac{\binom{38}{5}}{\binom{48}{5}} \] gives the likelihood of selecting only sharp blades, which is necessary for calculating other probabilities.

Independent Events

In probability theory, when we say two events are independent, it means the occurrence of one does not affect the probability of the other occurring. This is a pivotal assumption for this exercise since each day's evaluation is considered separate from the next.
This principle is crucial because it means the outcome of evaluating the blades on any given day doesn't influence the results on subsequent days.

• **Understanding Independence**: If Plate 1 is dull on day 1, it doesn't change the fact that Plate 2 might or might not be dull on day 2.
• **Probability Calculation**: To compute these independent probabilities, use the multiplication rule \[ P(A \text{ and } B) = P(A) \times P(B) \] which states that if events \( A \) and \( B \) are independent, their joint probability is the product of their individual probabilities.
• **Example**: In our problem, the probability the assembly isn't replaced until the third day involves independent probabilities over three days: \[ (1 - P_1(\text{dull})) \times (1 - P_2(\text{dull})) \times P_3(\text{dull}) \] showing that each day's evaluation is an independent event.