Question

Let \(y\) denote the number of broken eggs in a randomly selected carton of one dozen eggs. Suppose that the probability distribution of \(y\) is as follows: $$ \begin{array}{lrrrrl} y & 0 & 1 & 2 & 3 & 4 \\ p(y) & .65 & .20 & .10 & .04 & ? \end{array} $$ a. Only \(y\) values of \(0,1,2,3\), and 4 have positive probabilities. What is \(p(4)\) ? b. How would you interpret \(p(1)=.20 ?\) 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. \(p(4) = .01\), b. The probability that a randomly selected carton of eggs has exactly one broken egg is .20 or 20%. c. \(P(y ≤ 2) = .95\) or 95%, d. \(P(y < 2) = .85\) or 85%, this is lower because it doesn't include p(2). e. The probability that the carton contains exactly 10 unbroken eggs is 0.10 or 10%, f. The probability that at least 10 eggs are unbroken is .95 or 95%.

Step-by-step solution

Calculate p(4)

The probabilities of all possible outcomes must add up to 1. Since the probabilities for y = 0, 1, 2, and 3 are provided, the remaining probability for y = 4 can be calculated by subtracting the sum of these probabilities from 1. Hence, \(p(4) = 1 - (.65 + .20 + .10 + .04) = .01\).

Interpret p(1)=.20

Probability of an event is a measure of the likelihood of that event happening. Given p(1)=.20, it can be interpreted as the probability that a randomly selected carton of eggs has exactly one broken egg is .20 or 20%.

Calculate P(y ≤ 2)

This is the probability that the carton contains at most two broken eggs. It involves adding the probabilities of y values from 0 up to 2. So, \(P(y ≤ 2) = p(0) + p(1) + p(2) = .65 + .20 + .10 = .95 or 95%.\)

Calculate P(y < 2)

This is the probability that the carton contains fewer than two broken eggs. It involves adding the probabilities of y values less than 2. So, \(P(y < 2) = p(0) + p(1) = .65 + .20 = .85 or 85%. The reason this is smaller than the probability in Part (c) is due to not including p(2).\)

Calculate the probability that the carton contains exactly 10 unbroken eggs

If the carton contains exactly 10 unbroken eggs, it means it has exactly 2 broken eggs. Hence, this is equivalent to p(y=2), which is 0.10 or 10%.

Calculate the probability that at least 10 eggs are unbroken

If the carton contains at least 10 unbroken eggs, it means it contains 2,1, or 0 broken eggs. Hence, this is equivalent to P(y ≤ 2), which was calculated in Step 3 as .95 or 95%.