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Sick Leave The probability that a given worker at Dyno Nutrition will call in sick on a Monday is \(0.04 .\) The packaging department has eight workers. What is the probability that two or more packaging workers will call in sick next Monday?

Short Answer

Expert verified
The probability is 0.038.

Step by step solution

01

Identify Distribution Type

This problem involves calculating the probability of a certain number of successes in several trials, where each trial can have an outcome of either success (worker calls in sick) or failure (worker does not call in sick). Therefore, we use the Binomial distribution.
02

Determine Parameters

For the Binomial distribution, we need to determine the number of trials and the probability of success in each trial. Here, the number of workers (trials) is 8, and the probability that a worker calls in sick (success) is 0.04.
03

Express Probability Scenario

We need to find the probability that two or more workers call in sick. This is the complement of the probability that fewer than two (0 or 1) workers call in sick. Thus, we compute: \[ P(X \geq 2) = 1 - P(X < 2) = 1 - \left(P(X = 0) + P(X = 1)\right) \]
04

Calculate Probability for Zero Workers Sick

Use the binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]. For zero workers sick (\(k=0\)), compute: \[ P(X=0) = \binom{8}{0} (0.04)^0 (0.96)^8 = 1 \cdot 1 \cdot 0.96^8 \approx 0.721 \].
05

Calculate Probability for One Worker Sick

Again, using the binomial probability formula for \( k = 1 \):\[ P(X=1) = \binom{8}{1} (0.04)^1 (0.96)^7 = 8 \cdot 0.04 \cdot 0.96^7 \approx 0.241 \].
06

Compute Complementary Event

Now, sum the probabilities for zero and one worker:\[ P(X < 2) = P(X=0) + P(X=1) = 0.721 + 0.241 = 0.962 \].
07

Find Desired Probability

Use the complementary probability to find the probability of the event we are trying to determine:\[ P(X \geq 2) = 1 - P(X < 2) = 1 - 0.962 = 0.038 \].

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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 a branch of mathematics that helps us understand how likely an event is to occur. It's like a numerical language of uncertainty. Here, we care about finding probabilities in a specific context—the likelihood of certain workers being absent due to sickness.
For a specific context like a sick day, understanding the probability of zero, one, or more workers calling in sick is crucial. This sort of problem can help businesses manage their staffing effectively by anticipating possible absences.
The solution involves complement probability, a helpful method to determine our result indirectly by first finding what's unwelcome (zero or one sick day) and subtracting it from the total. This process allows calculation flexibility, making it an invaluable tool in probability theory.
Statistical Methods
Statistical methods form the foundation of analyzing and interpreting data. In this case, we used a Binomial distribution, which is one of the essential tools in statistics for dealing with two-outcome scenarios.
The Binomial distribution provides the probability of a given number of successes (workers calling in sick) in a fixed number of trials (number of workers), with a consistent probability of success (risk of calling in sick). Each worker represents an identical trial with two possible outcomes: sick or not sick.
When applying statistical methods, calculating probabilities requires detailed steps such as identifying the statistical model (like Binomial distribution), determining parameters, and applying probability formulas. Statistics thus acts like a compass, guiding us through interpreting seemingly chaotic data into meaningful insights.
Combinatorics
Combinatorics is a mathematical technique focusing on counting and arrangement possibilities. When determining the binomial probabilities, combinations come into play using the Binomial Coefficient \(\binom{n}{k}\), which counts the ways to choose \(k\) successes (sick workers) out of \(n\) trials (total workers).
In our problem, this coefficient calculates different ways sick days could happen among eight workers. For instance, for one worker to be sick out of eight, the combination counts how many possible single-outcome selections exist, helping to compute the exact probability of that occurrence.
Combinatorics simplifies complex decision-making scenarios, allowing businesses or fields that rely on probability and statistics to logically organize information and draw effective conclusions.

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Most popular questions from this chapter

Why Is \(\left(x^{n}\right)\) the Same as \(C(n, r) ?\) This exercise explains why the binomial coefficients \(\left(\begin{array}{c}{n} \\\ {r}\end{array}\right)\) that appear in the expansion of \((x+y)^{n}\) are the same as \(C(n, r),\) the number of ways of choosing \(r\) objects from \(n\) objects. First, note that expanding a binomial using only the Distributive Property gives $$ \begin{aligned}(x+y)^{2} &=(x+y)(x+y) \\ &=(x+y) x+(x+y) y \\ &=x x+x y+y x+y y \\\\(x+y)^{3} &=(x+y)(x x+x y+y x+y y) \\ &=x x x+x x y+x y x+x y y+y x x \\ &+y x y+y y x+y y y \end{aligned} $$ (a) Expand \((x+y)^{5}\) using only the Distributive Property. (b) Write all the terms that represent \(x^{2} y^{3}\) together. These are all the terms that contain two \(x^{\prime}\) s and three \(y^{\prime} s .\) (c) Note that the two \(x\) 's appear in all possible positions. Conclude that the number of terms that represent \(x^{2} y^{3}\) is \(C(5,2) .\) (d) In general, explain why \((r)\) in the Binomial Theorem is the same as \(C(n, r) .\)

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