/*! 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 14 At a certain junior high school,... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

At a certain junior high school, two-thirds of the students have at least one tooth cavity. A dental survey is made of the students What is the probability that the first student to have a cavity is the third student examined?

Short Answer

Expert verified
The probability is \(\frac{1}{27}\).

Step by step solution

01

Understand the Problem

Determine what the problem is asking for, specifically the probability that the first student to have a cavity is the third student examined.
02

Define the Probability

Let the probability of a student having a cavity be \( P(C) = \frac{2}{3} \) and the probability of a student not having a cavity be \( P(C^{'} ) = \frac{1}{3} \).
03

Set Up the Required Probability

We need to find the probability that the first student to have a cavity is the third student examined. This means the first two students do not have a cavity and the third one does. So, the probability is \( P(C^{'} ) \times P(C^{'}) \times P(C) \).
04

Calculate the Probability

Substitute the probabilities \( P(C^{'} ) = \frac{1}{3} \) and \( P(C) = \frac{2}{3} \) into the formula. Thus, \[ P(\text{First cavity on third student}) = \frac{1}{3} \times \frac{1}{3} \times \frac{2}{3} = \frac{1}{27} \].

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.

Probability Theory
Probability theory is the branch of mathematics that deals with the likelihood of different outcomes. It's about predicting future events based on known possibilities. For example, in our junior high school dental survey, we want to know the chance that the first student with a cavity is the third one examined.
The probability of an event happening is given by a fraction where:
  • The numerator represents the number of successful outcomes
  • The denominator represents the total number of possible outcomes
In our exercise, the probability that a student has a cavity is \(\frac{2}{3}\). The probability of a student not having a cavity is \(\frac{1}{3}\).
When dealing with multiple events, we need to multiply the probabilities of each individual event occurring.
Binomial Probability
The binomial probability deals with scenarios where there are exactly two possible outcomes, like success or failure. In our dental survey exercise, each student either has a cavity or doesn't. We can use the binomial probability formula here.
The general formula for binomial probability is given by: \[ P(X = k) = {n \choose k} \cdot p^k \cdot (1-p)^{n-k} \] where:
  • \(n\) is the total number of trials (or students)
  • \(k\) is the number of successful trials (students with cavities)
  • \(p\) is the probability of a success (cavity)
In our case, we're interested in the specific order: the first student having a cavity on the third try. This isn't a standard binomial problem since it requires a specific sequence of events. Instead, we calculate it by sequentially multiplying individual probabilities: \[ P(C^{'} ) \times P(C^{'} ) \times P(C) = \frac{1}{3} \times \frac{1}{3} \times \frac{2}{3} = \frac{1}{27} \]
Basic Statistics
Basic statistics helps us understand and interpret data. In the context of our problem, we use several fundamental concepts:
  • Mean: The average of a set of numbers, not directly relevant to our probability question but important in general data analysis.
  • Median: The middle value in a dataset when arranged in ascending order.
  • Mode: The most frequently occurring value in a dataset.
For our junior high school dental survey, we're particularly interested in probability, a core part of statistics. Understanding the likelihood of an event — like finding the first cavity on the third student examined — helps us make informed predictions. By breaking the problem down into steps and applying basic statistical and probability concepts, we determine that the probability is \( \frac{1}{27} \).
Statistics is not only about complex calculations but also about logical and critical thinking to solve real-life problems.

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

Compute the cumulative distribution function corresponding to the density function \(f(x)=\frac{1}{2}(3-x), 1 \leq x \leq 3.\)

The amount of time required to serve a customer at a bank has an exponential density function with a mean of 3 minutes. Find the probability that a customer is served in less than 2 minutes.

The number of phone calls coming into a telephone switchboard during each minute was recorded during an entire hour. During 30 of the 1-minute intervals there were no calls, during 20 intervals there was one call, and during 10 intervals there were two calls. A 1-minute interval is to be selected at random and the number of calls noted. Let \(X\) be the outcome. Then, \(X\) is a random variable taking on the values \(0,1,\) and \(2.\) (a) Write out a probability table for \(X.\) (b) Compute \(E(X).\) (c) Interpret \(E(X).\)

Illustrate a technique from statistics (called the method of maximum likelihood) that estimates a parameter for a probability distribution. In a production process, a box of fuses is examined and found to contain two defective fuses. Suppose that the probability of having two defective fuses in a box selected at random is \(\left(\lambda^{2} / 2\right) e^{-\lambda}\) for some \(\lambda\). Take first and second derivatives to determine the value of \(\lambda\) for which the probability has its maximum value.

During a certain part of the day, the time between arrivals of automobiles at the tollgate on a turnpike is an exponential random variable with an expected value of 20 seconds. Find the probability that the time between successive arrivals is more than 60 seconds.

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.