/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 89 Marriage Anniversaries Suppose a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Marriage Anniversaries Suppose all the months of the year are equally likely as marriage anniversaries. Glen and Shahid are two randomly selected married males (unrelated). a. What is the probability that they were both married in August? b. What is the probability that Glen OR Shahid was married in August? Hint: The answer is not \(2 / 12\) or \(1 / 6\). Refer to Guided Exercise \(5.25\) if you need help.

Short Answer

Expert verified
The probability that both were married in August is \(1 / 144\). The probability that either Glen or Shahid was married in August is \(23 / 144\).

Step by step solution

01

Calculate the probability that both were married in August

Since this is about two independent events, we can multiply the probabilities together. Every month probability that a man was married in is \(1 / 12\) since we assume all months are equally likely. So, the probability for both men being married in August is \( (1/12) * (1/12) = 1 / 144 \).
02

Calculate the probability that either of them was married in August

The probability that either Glen or Shahid was married in August is the sum of their individual probabilities minus the probability that both events happen as this was counted twice. Using the calculation from step 1, this equals to \( 2 * (1 / 12) - (1 / 144) = 1 / 6 - 1 / 144 = 23 / 144 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Independent Events
When we say two events are independent in statistics, we mean the occurrence of one event does not affect the outcome of another event. For example, flipping a coin and rolling a die are independent events; the result of the coin flip does not influence the die roll.

In the marriage anniversaries problem, Glen and Shahid's anniversaries are independent events. Glen's wedding month does not impact Shahid’s, and vice versa. Understanding this concept is crucial because it allows us to perform calculations like multiplying their probabilities to find the likelihood of both events occurring. To be more specific, if the probability of Glen being married in August is \(1/12\) and the same applies to Shahid, the probability of both being married in August is \(1/12 * 1/12\), which results in \(1/144\).
Probability Calculations
Probability calculations can sometimes be more nuanced than one might expect. For instance, calculating the probability that either Glen or Shahid was married in August does not simply mean adding their individual probabilities.

This is because when calculating the probability of 'either this or that' events, one must also subtract the probability of the event happening to both (since those instances have been counted twice). So the correct approach is to add the probabilities of each event occurring individually and then subtract the probability of them both occurring together. With numbers, this becomes \(2 * (1/12)\) minus the probability we've already determined they both get married in August \(1/144\), leaving us with \(23/144\). Remember, when working with probabilities, it’s essential to account for the overlap to avoid double-counting.
Combinatorics
Combinatorics is a branch of mathematics dealing with counting, combination, and permutation of sets of elements, and it plays a pivotal role in probability calculations. It gives us tools to count the number of possible outcomes and arrangements, which is fundamental in determining the probabilities of different events.

In simpler scenarios like the marriage anniversaries problem, we may not see the complexities of combinatorics at play, but the principles are there. Each man has 12 options (months), and when we calculate probabilities, we’re implicitly using the basic counting principle—there are 12 possible options for one man (12 individual months) and, thus, \(12 * 12 = 144\) possible combinations for two.

Simple yet Powerful

These principles can extend to more complex situations where the number of combinations becomes less apparent, and the formulas and techniques of combinatorics become invaluable to solving probability problems. The ability to count the correct number of desirable outcomes versus the total possible outcomes is what allows us to calculate probabilities accurately.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Recidivism (Example 16) Norway's recidivism rate is one of the lowest in the world at \(20 \%\). This means that about \(20 \%\) of released prisoners end up back in prison (within three years). Suppose three randomly selected prisoners who have been released are studied. a. What is the probability that all three of them go back to prison? What assumptions must you make to calculate this? b. What is the probability that neither of them goes back to prison? c. What is the probability that at least two go back to prison?

GSS: Political Party The General Social Survey (GSS) is a survey done nearly every year at the University of Chicago. One survey, summarized in the table, asked each respondent to report her or his political party affiliation and whether she or he was liberal, moderate, or conservative. (Dem stands for Democrat, and Rep stands for Republican.) $$ \begin{array}{lcccc} & \text { Dem } & \text { Rep } & \text { Other } & \text { Total } \\ \hline \text { Liberal } & 306 & 26 & 198 & 530 \\ \hline \text { Moderate } & 279 & 134 & 322 & 735 \\ \hline \text { Conservative } & 104 & 309 & 180 & 593 \\ \hline \text { Total } & 689 & 469 & 700 & 1858 \end{array} $$ a. If one person is chosen randomly from the group, what is the probability that the person is liberal? b. If one person is chosen randomly from the group, what is the probability that the person is a Democrat?

Playing Cards (Example 2) There are four suits: clubs cards appear in each suit: ace, \(2,3,4,5,6,7,8,9,10\), jack, queen, king. The jack, queen, and king are called face cards because they have a drawing of a face on them. Diamonds and hearts are red, and clubs and spades are black. If you draw 1 card randomly from a standard 52 -card playing deck, what is the probability that it will be: a. A heart? d. A face card (jack, queen, or king)? b. A red card? e. A three? c. An ace?

Guessing on Balls Consider a bag containing five balls of different colors (green, blue, red, white, and yellow) for each of these questions. a, What is the probability of guessing the draw of a blue ball if a ball is to be drawn only once? b. What is the probability of guessing the draw of any other ball in one draw?

Independent Variables Use your general knowledge to label the following pairs of variables as independent or associated. Explain. a. For a sample of adults, gender and ring size. b. The outcome on rolls of two separate, fair dices.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.