/*! 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 17 What is the probability that a b... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

What is the probability that a baby will be born on a Friday OR a Saturday OR a Sunday if all the days of the week are equally likely as birthdays?

Short Answer

Expert verified
The probability of a baby being born on either Friday, Saturday, or Sunday is \( \frac{3}{7} \).

Step by step solution

01

Determine the probabilities of individual events

First, define the probability of a baby being born on a specific day. Since there are 7 days in a week and each day is equally likely, the probability for each individual day (whether it's Friday, Saturday, or Sunday) is \( \frac{1}{7} \)
02

Add the three probabilities together

Since it's asked for the probability of the baby being born on either Friday, Saturday, or Sunday, these are disjoint events (a baby cannot be born on two days at once), so the probabilities should just be added together. \( P(\text{Friday or Saturday or Sunday}) = P(\text{Friday}) + P(\text{Saturday}) + P(\text{Sunday}) = \frac{1}{7} + \frac{1}{7} + \frac{1}{7} \)
03

Simplify the sum

Finally, simplify the sum from Step 2. \( \frac{1}{7} + \frac{1}{7} + \frac{1}{7} = \frac{3}{7} \). So, the probability of a baby being born on either Friday, Saturday, or Sunday is \( \frac{3}{7} \)

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.

Equally Likely Events
In probability, equally likely events refer to outcomes that have the same chance of occurring. Imagine a scenario where you have a fair die. Rolling any of the six numbers (1 through 6) has an equal probability because the die is not biased. In our exercise, we're dealing with days of the week when a baby could be born. Since there are 7 days available, and assuming no external factors affecting the birth day, each day (Monday through Sunday) is equally probable. This gives each day a probability of \( \frac{1}{7} \).

When talking about equally likely events, it's important to affirm that each event must always have precisely the same chance of happening for them to be considered equal. Real-life situations often rely on this principle for simplifying complex problems.

Equally likely events make calculations more manageable because of their uniform probability, allowing us to directly use these values in further probability rules.
Disjoint Events
Disjoint events, also known as mutually exclusive events, are events that cannot occur simultaneously. In simpler terms, if one event happens, the other cannot happen at the same time. This concept is embodied in our exercise with the days of birth. A baby cannot be born on both Friday and Saturday simultaneously; these events are separate with no overlap.

Understanding disjoint events helps us simplify probability calculations. When events are disjoint, the total probability of any of these events occurring is just the sum of their individual probabilities. Always remember, for events to be considered disjoint, their occurrence must be mutually exclusive, meaning they do not affect each other and cannot happen at the same time.
Probability Addition Rule
The probability addition rule is a fundamental concept used to find the probability of any one of several mutually exclusive events happening. This rule simplifies the addition of probabilities for disjoint events, which is exactly what we employed in our exercise.

For disjoint events, the probability addition rule states: if you want to find the probability of one event OR another event occurring, you simply add the probabilities of each event together. Mathematically, for two disjoint events \( A \) and \( B \), the probability \( P(A \text{ or } B) = P(A) + P(B) \).

In our exercise, this is evident as we looked for the probability of a baby being born on either Friday, Saturday, or Sunday. Using the rule, we added the probabilities directly: \( \frac{1}{7} + \frac{1}{7} + \frac{1}{7} = \frac{3}{7} \). This illustrates how the rule functions effectively for determining probabilities involving multiple disjoint events, making computations straightforward and intuitive.

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

A student flips a coin. 10 times and sees that it landed on tails 4 times. Based on this, the student says that the probability of getting a tail is \(40 \%\). Is the student referring to an empirical probability or a theoretical probability? Explain.

The sample space shows all possible sequences of child gender for a family with 3 children. The table is organized by the number of girls in the family. a. How many outcomes are in the sample space? b. If we assume all outcomes in the sample space are equally likely, find the probability of having the following numbers of girls in a family of 3 children: i. all 3 girls ii. no girls iii. exactly 2 girls

A Monopoly player claims that the probability of getting a 4 when rolling a six-sided die is \(1 / 6\) because the die is equally likely to land on any of the six sides. Is this an example of an empirical probability or a theoretical probability? Explain.

According to the National Center for Health Statistics, \(52 \%\) of U.S. households no longer have a landline and instead only have cell phone service. Suppose three U.S. households are selected at random. a. What is the probability that all three have only cell phone service? b. What is the probability that at least one has only cell phone service?

A driving exam consists of 30 multiplechoice questions. Each of the answers is either right or wrong. Suppose that the probability of making fewer than 7 mistakes is \(0.23\) and the probability of making from 7 to 15 mistakes is \(0.41\). Find the probability of making the following: a. 16 or more mistakes b. 7 or more mistakes c. At most 15 mistakes d. Which two of these three events are complementary? Explain.

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.