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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 soft-drink machine dispenses only regular Coke and Diet Coke. Sixty percent of all purchases from this machine are diet drinks. The machine currently has 10 cans of each type. If 15 customers want to purchase drinks before the machine is restocked, what is the probability that each of the 15 is able to purchase the type of drink desired?

Suppose that \(20 \%\) of all homeowners in an earthquakeprone area of California are insured against earthquake damage. Four homeowners are selected at random. Define the random variable \(x\) as the number among the four who have earthquake insurance. a. Find the probability distribution of \(x\). (Hint: Let \(S\) denote a homeowner who has insurance and \(\mathrm{F}\) one who does not. Then one possible outcome is SFSS, with probability (0.2)(0.8)(0.2)(0.2) and associated \(x\) value of 3 . There are 15 other outcomes.) b. What is the most likely value of \(x ?\) c. What is the probability that at least two of the four selected homeowners have earthquake insurance?

Thirty percent of all automobiles undergoing an emissions inspection at a certain inspection station fail the inspection. a. Among 15 randomly selected cars, what is the probability that at most 5 fail the inspection? b. Among 15 randomly selected cars, what is the probability that between 5 and 10 (inclusive) fail the inspection? c. Among 25 randomly selected cars, what is the mean value of the number that pass inspection, and what is the standard deviation? d. What is the probability that among 25 randomly selected cars, the number that pass is within 1 standard deviation of the mean value?

Airlines sometimes overbook flights. Suppose that for a plane with 100 seats, an airline takes 110 reservations. Define the random variable \(x\) as \(x=\) the number of people who actually show up for a sold-out flight on this plane From past experience, the probability distribution of \(x\) is given in the following table: $$ \begin{array}{|cc|} \hline \boldsymbol{x} & \boldsymbol{p}(\boldsymbol{x}) \\ \hline 95 & 0.05 \\ 96 & 0.10 \\ 97 & 0.12 \\ 98 & 0.14 \\ 99 & 0.24 \\ 100 & 0.17 \\ 101 & 0.06 \\ 102 & 0.04 \\ 103 & 0.03 \\ 104 & 0.02 \\ 105 & 0.01 \\ 106 & 0.005 \\ 107 & 0.005 \\ 108 & 0.005 \\ 109 & 0.0037 \\ 110 & 0.0013 \\ \hline \end{array} $$ a. What is the probability that the airline can accommodate everyone who shows up for the flight? b. What is the probability that not all passengers can be accommodated? c. If you are trying to get a seat on such a flight and you are number 1 on the standby list, what is the probability that you will be able to take the flight? What if you are number 3 ?

A contractor is required by a county planning department to submit anywhere from one to five forms (depending on the nature of the project) when applying for a building permit. Let \(y\) be the number of forms required of the next applicant. Suppose the probability that \(y\) forms are required is known to be proportional to \(y ;\) that is, \(p(y)=\) \(k y\) for \(y=1, \ldots, 5\) a. What is the value of \(k ?\) (Hint: \(\Sigma p(y)=1 .)\) b. What is the probability that at most three forms are required? c. What is the probability that between two and four forms (inclusive) are required?
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