/*! 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 204 The probability that any single ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The probability that any single driver will turn left at an intersection is .2. The left turn lane at this intersection has room for three vehicles. If the left turn lane is empty when the light turns red and five vehicles arrive at this intersection while the light is red, find the probability that the left turn lane will hold the vehicles of all of the drivers who want to turn left.

Short Answer

Expert verified
The probability is 0.99328.

Step by step solution

01

Define the Probability and Scenario

The probability that a single driver will turn left at the intersection is given as 0.2. We need to find the probability that one or more of those drivers fit in the left turn lane, which has a capacity for 3 vehicles, given a total of 5 vehicles arrive during the red light.
02

Clarify the Random Variable and Conditions

Let the random variable \(X\) be the number of drivers who want to turn left. \(X\) follows a binomial distribution \(X \sim \text{Binomial}(n=5, p=0.2)\), where \(n\) is the number of trials (vehicles arriving) and \(p\) is the probability of success (turning left for each trial). We want the probability that \(X \leq 3\).
03

Calculate Probabilities of Each Suitable Event

Calculate the probability that \(X = 0\), \(X = 1\), \(X = 2\), and \(X = 3\). Use the binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] Substitute \(n=5\) and \(p=0.2\) for each value of \(k\).
04

Probability Calculations

- \(P(X = 0) = \binom{5}{0} (0.2)^0 (0.8)^5 = 0.32768\)- \(P(X = 1) = \binom{5}{1} (0.2)^1 (0.8)^4 = 0.4096 \)- \(P(X = 2) = \binom{5}{2} (0.2)^2 (0.8)^3 = 0.2048\)- \(P(X = 3) = \binom{5}{3} (0.2)^3 (0.8)^2 = 0.0512\).
05

Calculate Total Probability

Add these probabilities to find the probability that \(X \leq 3\):\( P(X \leq 3) = 0.32768 + 0.4096 + 0.2048 + 0.0512 = 0.99328 \).

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 Calculation
In probability theory, calculating the likelihood of a particular event involves gathering relevant data and then using mathematical formulas to estimate the chances of that event occurring. Here, we are dealing with the probability that a certain number of drivers will choose to turn left at an intersection.

Once you have the basic data, such as the probability of a single event (in this case, a driver turning left), you can use these to assess compound events. For specific probabilities like a number of people turning left from a group, a structured approach ensures accuracy.

Make note that each step in the calculation builds on the prior steps. The calculations in each portion, like for zero drivers or multiple drivers wanting to turn left, must be determined separately and then aggregated correctly.

This calculation provides insight into how many drivers will want to turn left and whether the left turn lane can accommodate them. By summing up the probabilities for various outcomes, we can conclude with a total probability of the event where the left turn lane will hold those drivers.
Random Variables
Random variables are critical components in probability and statistics, providing a way to represent and quantify random phenomena.

Here, our random variable is the number of drivers wishing to turn left, represented as \(X\). In this scenario, \(X\) is modeled as a binomial random variable, reflecting the fact that the event of turning left is independent for each driver and has only two outcomes—either the driver turns left (success) or not (failure).

Random variables like \(X\) take on different values with specified probabilities. These values provide insight into how likely it is for the situation to fall within a certain range. This helps in predicting outcomes and assists in making informed decisions based on the calculated likelihoods.

Understanding this helps clarify why it is necessary to define a suitable random variable for any probability-related problem before starting with calculations.
Binomial Probability Formula
The binomial probability formula is fundamental for calculating probabilities in experiments or trials that follow a binomial distribution. In this case, the number of drivers turning left follows a binomial distribution, defined by parameters \(n\) and \(p\), where \(n\) is the number of trials (5 vehicles arriving), and \(p\) is the probability of success (0.2 probability of turning left).

The formula is given by: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]This represents the probability of having exactly \(k\) successes in \(n\) trials.

To use this formula, substitute the given values for each \(k\) scenario you are interested in. Calculations for scenarios where \(X = 0, 1, 2, 3\) turn into individual probabilities that must be summed up to determine \(P(X \leq 3)\), the probability that at most 3 drivers wish to turn left.

Understanding this formula helps you not only compute the necessary probabilities but also provides insights into the behavior of drivers at intersections and other similar everyday decisions.

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

Would you rather take a multiple-choice test or a full-recall test? If you have absolutely no knowledge of the test material, you will score zero on a full-recall test. However, if you are given 5 choices for each multiple-choice question, you have at least one chance in five of guessing each correct answer! Suppose that a multiple-choice exam contains 100 questions, each with 5 possible answers, and guess the answer to each of the questions. a. What is the expected value of the number \(Y\) of questions that will be correctly answered? b. Find the standard deviation of \(Y\). c. Calculate the intervals \(\mu \pm 2 \sigma\) and \(\mu \pm 3 \sigma\) d. If the results of the exam are curved so that 50 correct answers is a passing score, are you likely to receive a passing score? Explain.

The number of knots in a particular type of wood has a Poisson distribution with an average of 1.5 knots in 10 cubic feet of the wood. Find the probability that a 10 -cubic-foot block of the wood has at most 1 knot.

Suppose that \(Y\) is a discrete random variable with mean \(\mu\) and variance \(\sigma^{2}\) and let \(W=2 Y\) a. Do you expect the mean of \(W\) to be larger than, smaller than, or equal to \(\mu=E(Y) ?\) Why? b. Use Theorem 3.4 to express \(E(W)=E(2 Y)\) in terms of \(\mu=E(Y) .\) Does this result agree with your answer to part (a)? c. Recalling that the variance is a measure of spread or dispersion, do you expect the variance of \(W\) to be larger than, smaller than, or equal to \(\sigma^{2}=V(Y) ?\) Why? d. Use Definition 3.5 and the result in part (b) to show that $$V(W)=E\left\\{[W-E(W)]^{2}\right\\}=E\left[4(Y \mu)^{2}\right]=4 \sigma^{2}$$; that is, \(W=2 Y\) has variance four times that of \(Y\).

In responding to a survey question on a sensitive topic (such as "Have you ever tried marijuana?"), many people prefer not to respond in the affirmative. Suppose that \(80 \%\) of the population have not tried marijuana and all of those individuals will truthfully answer no to your question. The remaining \(20 \%\) of the population have tried marijuana and \(70 \%\) of those individuals will lie. Derive the probability distribution of \(Y\), the number of people you would need to question in order to obtain a single affirmative response.

The sizes of animal populations are often estimated by using a capture-tag- recapture method. In this method \(k\) animals are captured, tagged, and then released into the population. Some time later n animals are captured, and \(Y\), the number of tagged animals among the \(n\), is noted. The probabilities associated with \(Y\) are a function of \(N\), the number of animals in the population, so the observed value of \(Y\) contains information on this unknown \(N\). Suppose that \(k=4\) animals are tagged and then released. A sample of \(n=3\) animals is then selected at random from the same population. Find \(P(Y=1)\) as a function of \(N\). What value of \(N\) will maximize \(P(Y=1) ?\)

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.