91Ó°ÊÓ

Suppose \(y=\) the number of broken eggs in a randomly selected carton of one dozen eggs. The probability distribution of \(y\) is as follows: $$ \begin{array}{lccccl} y & 0 & 1 & 2 & 3 & 4 \\ p(y) & 0.65 & 0.20 & 0.10 & 0.04 & ? \end{array} $$ a. Only \(y\) values of \(0,1,2,3,\) and 4 have probabilities greater than 0 . What is \(p(4)\) ? b. How would you interpret \(p(1)=0.20 ?\) c. Calculate \(P(y \leq 2)\), the probability that the carton contains at most two broken eggs, and interpret this probability. d. Calculate \(P(y<2),\) the probability that the carton contains fewer than two broken eggs. Why is this smaller than the probability in Part (c)? e. What is the probability that the carton contains exactly 10 unbroken eggs? f. What is the probability that at least 10 eggs are unbroken?

Step by step solution

01

Find \(p(4)\)

We know that a valid probability distribution over a discrete random variable should add up to 1. So we sum the known probabilities and subtract this total from 1: \( p(4) = 1 - (p(0) + p(1) + p(2) + p(3)) = 1 - (0.65 + 0.20 + 0.10 + 0.04) = 0.01.

02

Interpret \(p(1)=0.20\)

\(p(1)=0.20\) means there is a 20% chance of finding exactly one broken egg in a randomly chosen carton.

03

Calculate \(P(y \leq 2)\)

\(P(y \leq 2)\) is the probability that the carton contains at most two broken eggs. We find this by adding the probabilities of \(y=0\), \(y=1\), and \(y=2\), thus: \(P(y \leq 2) = p(0) + p(1) + p(2) = 0.65 + 0.20 + 0.10 = 0.95.\

04

Calculate \(P(y

\(P(y<2)\) is the sum of the probabilities of \(y=0\) and \(y=1\), thus: \(P(y<2) = p(0) + p(1) = 0.65 + 0.20 = 0.85. This probability is less than the probability in Part (c) as \(P(y<2)\) does not include the event that there are two broken eggs.

05

Probability for exactly 10 unbroken eggs

Having exactly 10 unbroken eggs in a dozen corresponds to having exactly 2 broken eggs. Therefore, \(P(\text{10 unbroken eggs}) = P(y = 2) = 0.10.\

06

Probability for at least 10 unbroken eggs

Having at least 10 unbroken eggs means that we have 0, 1, or 2 broken eggs. So we add the probabilities of these cases: \(P(\text{at least 10 unbroken eggs}) = P(y = 0) + P(y = 1) + P(y = 2) = 0.65 + 0.20 + 0.10 = 0.95.\

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Key Concepts

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

Understanding Discrete Random Variables

A discrete random variable is a type of variable that can take on a finite or countable number of possible outcomes. In our exercise, the discrete random variable is \( y \), which represents the number of broken eggs in a carton. This means \( y \) can take values such as 0, 1, 2, 3, and 4, but not any value in between these numbers. Discrete random variables are often used in scenarios where each outcome is distinct and countable, like flipping a coin or rolling a die.
Understanding these variables is essential for calculating probabilities because they help establish what outcomes we are considering. In this case, the outcomes are the potential number of broken eggs.

Probability Interpretation

Probability interpretation involves understanding what the probabilities assigned to different outcomes mean. Take the probability \( p(1)=0.20 \) as an example. This value tells us there is a 20% chance that exactly one egg in the carton is broken.

• Each probability specifies the likelihood of a particular result.
• These probabilities must add context to the scenario, indicating how likely each outcome is.[li]

With this interpretation, you can also compare scenarios, like understanding that \( p(0)=0.65 \) means there is a 65% chance that no eggs are broken, making it the most likely outcome.

Sum of Probabilities

The sum of probabilities for all possible outcomes of a discrete random variable must equal 1. This concept ensures that one of the listed outcomes must occur.

For example, to find \( p(4) \), we need to sum all the known probabilities and subtract from 1: \( p(4) = 1 - (0.65 + 0.20 + 0.10 + 0.04) = 0.01 \). This confirms that every listed outcome is accounted for, maintaining the total probability at 100%.

• Ensures completeness of the probability distribution.
• Verifies that all probabilities have been assigned correctly and logically.

Interpretation of Probabilities

Interpreting probabilities involves connecting numerical values to real-world meanings. Let's consider \( P(y \leq 2) = 0.95 \), which indicates that there's a 95% probability of having at most two broken eggs. This shows the high likelihood that the majority of the eggs are unbroken.

When calculating \( P(y<2) \), which is only 85%, we observe it is lower because it excludes the event where exactly two eggs are broken.

• When interpreting probabilities, context is essential.
• This understanding helps in predicting outcomes and making informed decisions.

Such interpretations can guide actions, such as quality control in egg packaging, by highlighting the most probable scenarios.