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Chapter 6: Problem 85

Suppose that \(5 \%\) of cereal boxes contain a prize and the other \(95 \%\) contain the message, "Sorry, try again." Consider the random variable \(x,\) where \(x=\) number of boxes purchased until a prize is found. a. What is the probability that at most two boxes must be purchased? b. What is the probability that exactly four boxes must be purchased? c. What is the probability that more than four boxes must be purchased?

Short Answer

Expert verified
a. The probability of at most two boxes needing to be purchased is \(P(X \leq 2) \approx 0.0975\). b. The probability that exactly four boxes must be purchased is \(P(X = 4) 鈮� 0.043\). c. The probability that more than four boxes must be purchased is \(P(X > 4) 鈮� 0.8144\).

Step by step solution

01

- Recalling the Geometric Distribution Formula

The formula for the geometric probability distribution is given by: \[P(X = k) = (1-p)^{(k-1)}p\] Here, \(X\) is the random variable representing the number of trials until the first success, \(k\) is the specific number of trials we want to find the probability for, and \(p\) is the probability of success. In this problem, we have \(p=0.05\) and \(1-p=0.95\).
02

- Calculate the probability of at most two boxes

For part a, we want to find the probability that at most two boxes must be purchased. This means we want to find: \[P(X \leq 2) = P(X=1) + P(X=2)\] Using the geometric distribution formula from Step 1, we have: \[P(X=1) = (1-0.05)^{(1-1)}0.05 = 1^{0}0.05 = 0.05\] \[P(X=2) = (1-0.05)^{(2-1)}0.05 = 0.95^{1}0.05 \approx 0.0475\] Therefore, the probability of at most two boxes needing to be purchased is: \[P(X \leq 2) 鈮� 0.05 + 0.0475 = 0.0975\]
03

- Calculate the probability of exactly four boxes

For part b, we want to find the probability that exactly four boxes must be purchased. This means we want to find: \[P(X = 4)\] Using the geometric distribution formula from Step 1, we have: \[P(X=4) = (1-0.05)^{(4-1)}0.05 = 0.95^{3}0.05 \approx 0.043\] Therefore, the probability that exactly four boxes must be purchased is: \[P(X = 4) 鈮� 0.043\]
04

- Calculate the probability of more than four boxes

For part c, we want to find the probability that more than four boxes must be purchased. This means we want to find: \[P(X > 4) = 1 - P(X \leq 4)\] Because \(X=1\), \(X=2\), \(X=3\), and \(X=4\) are the only possibilities less than or equal to 4, we have: \[P(X \leq 4) = P(X=1) + P(X=2) + P(X=3) + P(X=4)\] We already found \(P(X=1)\), \(P(X=2)\), and \(P(X=4)\) in previous steps. To find \(P(X=3)\), we can use the geometric distribution formula: \[P(X=3) = (1-0.05)^{(3-1)}0.05 = 0.95^{2}0.05 \approx 0.0451\] So, the probability of purchasing four or fewer boxes is: \[P(X \leq 4) 鈮� 0.05 + 0.0475 + 0.0451 + 0.043 = 0.1856\] Finally, we can find the probability of purchasing more than four boxes: \[P(X > 4) = 1 - P(X \leq 4) 鈮� 1 - 0.1856 = 0.8144\]

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Key Concepts

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

Probability of Success
Understanding the probability of success is crucial when working with geometric distributions. In layman's terms, it's like knowing the odds that you'll win a prize when you open a cereal box if only some of them have a surprise inside.

In the example with cereal boxes, we consider winning a prize as a 'success' and everything else a 'failure'. Here, since 5% of boxes have a prize, the probability of success, denoted as 'p', is 0.05 or 5%. This tiny number has a big role in our calculations. Every time you pick a box, you have a 5% chance of striking gold and finding that prize. If you keep getting 'Sorry, try again' messages, your chances don't suddenly get better 鈥� they stay at 5% for each new box.
Random Variable
In probability and statistics, a random variable is not just any number you think of; it's a way to represent the outcome of a 'random' process with numbers. It's like labeling each outcome of a game of chance with a score.

In our cereal box scenario, the random variable 'x' represents the number of cereal boxes you buy until you find a prize. So, if 'x' equals 1, that means you found a prize in the very first box. If 'x' is 3, you had to buy three boxes before getting lucky. This variable can take on any positive integer value, indicating the box number where your search for the prize ends. The key point here is that 'x' is not just any number; it鈥檚 the specific point in your sequence of tries where you finally win.
Probability Distribution
The term 'probability distribution' might sound intimidating, but it's simply a way to list all the possible outcomes of a random event and the likelihood of each. Think of it as a menu of all the potential results of your cereal box adventure, along with the chances of each happening.

In the context of geometric distribution, it features a peculiar characteristic 鈥� the memoryless property. This means regardless of how many 'Sorry' boxes you've already opened, your chances of opening a 'prize' box remain constant. It distributes the probability across an infinite number of possible outcomes, indicating that you could theoretically keep opening boxes forever until you find a prize. To calculate these probabilities, formulas come into play, as seen in the cereal box problem, to provide exact numbers for these chances, paving the way to a proper understanding of geometric distribution in practice.

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Most popular questions from this chapter

A pizza shop sells pizzas in four different sizes. The 1000 most recent orders for a single pizza resulted in the following proportions for the various sizes: $$ \begin{array}{lcccc} \text { Size } & 12 \text { in. } & 14 \text { in. } & 16 \text { in. } & 18 \text { in. } \\ \text { Proportion } & 0.20 & 0.25 & 0.50 & 0.05 \end{array} $$ With \(x=\) the size of a pizza in a single-pizza order, the given table is an approximation to the population distribution of \(x\). a. Write a few sentences describing what you would expect to see for pizza sizes over a long sequence of single-pizza orders. b. What is the approximate value of \(P(x<16)\) ? c. What is the approximate value of \(P(x \leq 16)\) ?

An appliance dealer sells three different models of freezers having 13.5,15.9 , and 19.1 cubic feet of storage space. Consider the random variable \(x=\) the amount of storage space purchased by the next customer to buy a freezer. Suppose that \(x\) has the following probability distribution: \(x\) \(\begin{array}{lll}13.5 & 15.9 & 19.1\end{array}\) \(p(x)\) \(\begin{array}{lll}0.2 & 0.5 & 0.3\end{array}\) a. Calculate the mean and standard deviation of \(x\). (Hint: See Example \(6.15 .\) ) b. Give an interpretation of the mean and standard deviation of \(x\) in the context of observing the outcomes of many purchases.

Consider the random variable \(y=\) the number of broken eggs in a randomly selected carton of one dozen eggs. Suppose the probability distribution of \(y\) is as follows: \(\begin{array}{cccccc}y & 0 & 1 & 2 & 3 & 4 \\ p(y) & 0.65 & 0.20 & 0.10 & 0.04 & ?\end{array}\) 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?

6.89 Suppose that \(20 \%\) of the 10,000 signatures on a certain recall petition are invalid. Would the number of invalid signatures in a sample of size 2000 have (approximately) a binomial distribution? Explain.

The probability distribution of \(x,\) the number of defective tires on a randomly selected automobile checked at a certain inspection station, is given in the following table: \(\begin{array}{lccccc}x & 0 & 1 & 2 & 3 & 4 \\ p(x) & 0.54 & 0.16 & 0.06 & 0.04 & 0.20\end{array}\) a. Calculate the mean value of \(x\). b. Interpret the mean value of \(x\) in the context of a long sequence of observations of number of defective tires. c. What is the probability that \(x\) exceeds its mean value? d. Calculate the standard deviation of \(x\).
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