Question

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?

Short answer

a. \(p(4) = 0.01\), b. There's a 20% chance of finding one broken egg. c. \(P(y \leq 2) = 0.95.\) It represents the probability of finding at most 2 broken eggs in the carton. d. \(P(y < 2) = 0.85.\) It excludes the situation of having 2 broken eggs, hence it is smaller. e. The probability is 0.10. f. The probability is 0.95.

Step-by-step solution

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.

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.

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.\

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.

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.\

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.\