/*! 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 10 Let \(X\) be the winnings of a g... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(X\) be the winnings of a gambler. Let \(p(i)=\) \(P(X=i)\) and suppose that $$\begin{array}{l} p(0)=1 / 3 ; p(1)=p(-1)=13 / 55 \\ p(2)=p(-2)=1 / 11 ; p(3)=p(-3)=1 / 165 \end{array}$$ Compute the conditional probability that the gambler wins \(i, i=1,2,3,\) given that he wins a positive amount.

Short Answer

Expert verified
The conditional probabilities of the gambler winning \(i\) given they win a positive amount are: $$ P(X=1|X>0) = \frac{39}{152}; \enspace P(X=2|X>0) = \frac{15}{152}; \enspace P(X=3|X>0) = \frac{1}{152} $$

Step by step solution

01

Calculate the total probability of winning a positive amount

To find the total probability of the gambler winning a positive amount, we need to add the probabilities of winning 1, 2, or 3, that is: \(p(1) + p(2) + p(3)\). We are given: \(p(1) = \frac{13}{55}, p(2) = \frac{1}{11},\) and \(p(3) = \frac{1}{165}\). Adding these probabilities together, we get: Total probability of winning a positive amount = \(p(1)+p(2)+p(3) = \frac{13}{55} + \frac{1}{11} + \frac{1}{165}\)
02

Simplify to find the total probability

To simplify this, we find the common denominator and add the fractions: Total probability = \(\frac{13*3*3 + 1*5*3 + 5*11}{55*3} = \frac{39*3 + 15 + 55}{165} = \frac{117 + 15 + 55}{165} = \frac{187}{165} = 1 - \frac{13}{165}\)
03

Calculate the conditional probabilities

Now that we have the total probability of winning a positive amount, we can find the conditional probability of the gambler winning \(i\) given that they win a positive amount using the following formula: \(P(A|B) = \frac{P(A \cap B)}{P(B)}\) In our context, A is the event that the gambler wins \(i\), and B is the event that they win a positive amount. We can rewrite the formula for each \(i = 1, 2, 3\): \(P(X=i|X>0) = \frac{P(X=i \cap X>0)}{P(X>0)}\) Since \(X=i \cap X>0\) is equivalent to \(X=i\) for \(i=1,2,3\), we can rewrite the formula as: \(P(X=i|X>0) = \frac{P(X=i)}{P(X>0)}\) Now, we can substitute the values of \(p(i)\) and \(P(X>0)\) to find the probabilities for \(i=1,2,3\): 1. For \(i=1\): \(P(X=1|X>0) = \frac{p(1)}{P(X>0)} = \frac{\frac{13}{55}}{1 - \frac{13}{165}}\) 2. For \(i=2\): \(P(X=2|X>0) = \frac{p(2)}{P(X>0)} = \frac{\frac{1}{11}}{1 - \frac{13}{165}}\) 3. For \(i=3\): \(P(X=3|X>0) = \frac{p(3)}{P(X>0)} = \frac{\frac{1}{165}}{1 - \frac{13}{165}}\)
04

Simplify the conditional probabilities

Now, we'll simplify the expressions to find the final conditional probabilities: 1. For \(i=1\): \(P(X=1|X>0) = \frac{\frac{13}{55}}{1 - \frac{13}{165}} = \frac{13*3}{165-13} = \frac{39}{152}\) 2. For \(i=2\): \(P(X=2|X>0) = \frac{\frac{1}{11}}{1 - \frac{13}{165}} = \frac{15}{165-13} = \frac{15}{152}\) 3. For \(i=3\): \(P(X=3|X>0) = \frac{\frac{1}{165}}{1 - \frac{13}{165}} = \frac{1}{165-13} = \frac{1}{152}\) The conditional probabilities of the gambler winning \(i\) given they win a positive amount are: $$ P(X=1|X>0) = \frac{39}{152}; \enspace P(X=2|X>0) = \frac{15}{152}; \enspace P(X=3|X>0) = \frac{1}{152} $$

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Key Concepts

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

Probability Theory
Probability theory is a fundamental aspect of mathematics that deals with quantifying the likelihood of events. It is a branch of mathematics concerned with the analysis of random phenomena. The central objects of probability theory are random variables, stochastic processes, and events. Probability is quantified as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty.

In the context of the exercise, we're examining a simple probability case involving a gambler's winnings, represented by a discrete random variable, which can take values indicating losses or gains. The probabilities for each of these values are provided and are used to calculate conditional probabilities, a fundamental concept to understand the chances of an event occurring given that another event has already occurred. Understanding these fundamentals allows for the analysis of more complex stochastic situations in various fields such as finance, science, and engineering.
Positive Amount Probability
The 'positive amount probability' pertains to the likelihood of an outcome being more than zero. In gambling terms, it's the probability that the gambler comes out with a gain instead of a loss or breaking even. This is computed by summing the individual probabilities of every positive outcome. A common mistake made by students when assessing positive amount probability is failing to exclude the non-positive outcomes from the calculation, which would result in an incorrect probability.

The total probability of winning a positive amount in the exercise was calculated by summing up the probabilities of the gambler winning 1, 2, or 3 units. It's vital that students understand the inclusion-exclusion principle here to avoid overcounting when dealing with events that can overlap, which was not a concern in this particular exercise as the winning amounts are distinct outcomes.
Gambler's Winnings Probabilities
Gambler's winnings probabilities focus on the chance of various outcomes regarding gains or losses in scenarios such as games of chance. It's crucial to differentiate between the probability of winning any amount and the probability of winning a specific amount when dealing with conditional probabilities.

In the exercise provided, once the total probability of winning a positive amount is calculated, conditional probabilities must be computed. This involves the understanding that the probability of winning each specific positive amount ('event A') given that the gambler wins something positive ('event B') can be determined using the conditional probability formula:
\(P(A|B) = \frac{P(A \cap B)}{P(B)}\).
For students, it is important to comprehend that in \(P(A|B)\), event 'A' happening is only considered within the context of 'B' having occurred. Additionally, since the events are mutually exclusive (winning 1, 2, or 3 are separate outcomes), the intersection \(P(A \cap B)\) simplifies to \(P(A)\) for each case of winning amount 'i'. The focus on the positive winning probabilities eliminates the possibility of losses from our calculation framework, sharpening the perspective on calculating the gambler's fortune when it's already known they've won.

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