/*! 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 62 Suppose that the probability tha... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose that the probability that a head appears when a coin is tossed is \(p\) and the probability that a tail occurs is \(q=1-p .\) Person A tosses the coin until the first head appears and stops. Person B does likewise. The results obtained by persons \(A\) and \(B\) are assumed to be independent. What is the probability that \(A\) and \(B\) stop on exactly the same number toss?

Short Answer

Expert verified
The probability is \(\frac{p}{2-p}\).

Step by step solution

01

Understanding Geometric Distribution

The probability that person A stops on the \(n\)-th toss due to getting the first head is a geometric distribution problem. The probability of getting a tail first \((n-1)\) times and a head on the \(n\)-th toss is given by \(q^{n-1}p\). Thus, the probability that person A stops at the \(n\)-th toss is \(P(A_n) = q^{n-1}p\).
02

Calculate Probability for Person B

Similarly, person B will stop on the \(n\)-th toss with the same probability distribution since the events are independent and identical. Thus, the probability that person B stops at the \(n\)-th toss is \(P(B_n) = q^{n-1}p\).
03

Find Joint Probability of Both Stopping on the Same Toss

Since the results are independent, to find the probability that both A and B stop on the same toss, we multiply their individual probabilities: \(P(A_n \text{ and } B_n) = P(A_n) \times P(B_n) = (q^{n-1}p) \times (q^{n-1}p) = q^{2(n-1)}p^2\).
04

Sum Over All Possible Toss Outcomes

To find the total probability that A and B stop on the same number toss, we sum over all possible values of \(n\) (from 1 to infinity): \[P(\text{same n}) = \sum_{n=1}^{\infty} q^{2(n-1)}p^2 = p^2 \sum_{n=1}^{\infty} q^{2(n-1)}.\] This simplifies to a geometric series whose sum is \(\frac{1}{1-q^2}\) (since \(|q^2| < 1\)).
05

Evaluate the Geometric Series

The sum of the geometric series is \(\sum_{n=1}^{\infty} q^{2(n-1)} = \frac{1}{1-q^2}\). Therefore, the probability that A and B both stop on exactly the same number toss is \[P(\text{same n}) = p^2 \times \frac{1}{1-q^2} = \frac{p^2}{1-q^2}.\] Since \(q = 1-p\), we substitute to get \(1 - q^2 = 1 - (1-p)^2 = p(2-p)\).
06

Final Probability Calculation

Substituting back, we find that the probability that person A and B stop on the same toss is given by \[P(\text{same n}) = \frac{p^2}{p(2-p)} = \frac{p}{2-p}.\]
07

Conclusion

Therefore, the probability that A and B stop on exactly the same number toss is \(\frac{p}{2-p}\).

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 mathematical framework that allows us to quantify uncertainty by modeling random events. It helps us calculate how likely something is to happen. In the context of the coin toss problem, probability theory is used to determine the likelihood of specific outcomes, such as stopping at a certain toss.
A fundamental component of probability theory is the concept of random variables, which in our case, represent the number of tosses needed before getting a head. This leads directly to the geometric distribution, as each toss is an independent trial with two possible outcomes: head or tail.
Probability theory provides the tools to compute the likelihood of compound events, like the chance of both A and B stopping after the same number of tosses, by using rules for independent events. Understanding these principles can aid in solving complex problems systematically.
Independent Events
Independent events are outcomes that do not affect each other's occurrence. In the coin toss scenario, each toss by person A and person B is independent, meaning the result of person A's tosses does not influence person B's results. This is essential, as it allows us to treat each trial individually in probability calculations.
Essentially, the probability of two independent events both happening (such as A and B each stopping on the 4th toss) is the product of their individual probabilities. Since the event that person A gets a head on a specific toss does not alter the likelihood of person B's outcome, it simplifies the calculations and leads to accurate predictions based on initial assumptions of independence. This makes probability computations in stochastic processes straightforward.
Geometric Series
A geometric series is a series of terms that have a constant ratio between successive terms. In our exercise, we use a geometric series to mathematically model the cumulative probability across infinite trials. The sum of a geometric series can be represented as \( \frac{1}{1-r} \) when \( |r| < 1 \). This is crucial for solving our problem, where \( r = q^2 \), representing the probability of both A and B stopping after the same number of tosses for infinite potential values.
Here, the geometric series form simplifies the computation among potentially infinitely large values of \( n \), emphasizing the relationship between the series' sum and the probability of the combined occurrences. This allows us to distill complex and numerous potential outcomes into a palatable mathematical form, aiding in the computation of total probability.
Coin Toss Problem
The coin toss problem is a classic example used to illustrate the principles of probability, independence, and geometric series in action. In this scenario, two people toss coins until a head appears, representing each trial's independent event. The challenge is to determine the probability that both individuals experience the same number of tosses before getting a head.
By examining this problem, we use the concept of geometric distribution to model when a person stops tossing. Moreover, the problem utilizes probability theory and independence principles by multiplying individual probabilities to find the joint probability. Lastly, the solution's elegance lies in harnessing geometric series to sum probabilities over an infinite sequence of trials, offering an efficient means to the final probability \( \frac{p}{2-p} \). Thus, the coin toss problem distills complex probabilistic concepts into practical application.

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 target for a bomb is in the center of a circle with radius of 1 mile. A bomb falls at a randomly selected point inside that circle. If the bomb destroys everything within \(1 / 2\) mile of its landing point, what is the probability that the target is destroyed?

Contracts for two construction jobs are randomly assigned to one or more of three firms, A. B. and C. Let Y 1 denote the number of contracts assigned to firm A and Y 2 the number of contracts assigned to firm B. Recall that each firm can receive \(0.1 .\) or 2 contracts. a. Find the joint probability function for \(Y_{1}\) and \(Y_{2}\). b. Find \(F(1,0)\).

Two telephone calls come into a switchboard at random times in a fixed one- hour period. Assume that the calls are made independently of one another. What is the probability that the calls are made a. in the first half hour? b. within five minutes of each other?

A box contains \(N_{1}\) white balls, \(N_{2}\) black balls, and \(N_{3}\) red balls \(\left(N_{1}+N_{2}+N_{3}=N\right) .\) A random sample of \(n\) balls is selected from the box (without replacement). Let \(Y_{1}, Y_{2},\) and \(Y_{3}\) denote the number of white, black, and red balls, respectively, observed in the sample. Find the correlation coefficient for \(\left.Y_{1} \text { and } Y_{2} . \text { (Let } p_{i}=N_{i} / N, \text { for } i=1,2,3 .\right)\)

The joint density function of \(Y_{1}\) and \(Y_{2}\) is given by $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} 30 y_{1} y_{2}^{2}, & y_{1}-1 \leq y_{2} \leq 1-y_{1}, 0 \leq y_{1} \leq 1 \\ 0, & \text { elsewhere } \end{array}\right.$$ a. Find \(F(1 / 2,1 / 2)\). b. Find \(F(1 / 2,2)\). c. Find \(P\left(Y_{1}>Y_{2}\right)\).

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.