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In a poker hand, John has a very strong hand and bets 5 dollars. The probability that Mary has a better hand is .04. If Mary had a better hand she would raise with probability \(.9,\) but with a poorer hand she would only raise with probability .1. If Mary raises, what is the probability that she has a better hand than John does?

Short Answer

Expert verified
The probability that Mary has a better hand if she raises is approximately 0.2727.

Step by step solution

01

Define the Events

Let's define the relevant events: Let \( B \) be the event that Mary has a better hand, and \( R \) be the event that Mary raises.
02

Write Down Known Probabilities

We are given the following probabilities: \( P(B) = 0.04 \), \( P(R \mid B) = 0.9 \), and \( P(R \mid B^c) = 0.1 \).
03

Use Bayes' Theorem

We need to find the probability that Mary has a better hand given that she raises, i.e., \( P(B \mid R) \). According to Bayes' theorem: \[ P(B \mid R) = \frac{P(R \mid B)P(B)}{P(R)} \]
04

Calculate the Total Probability of Raising

The total probability that Mary raises, \( P(R) \), can be found by considering both scenarios: \[ P(R) = P(R \mid B)P(B) + P(R \mid B^c)P(B^c) \] Substituting the known probabilities, \[ P(R) = (0.9)(0.04) + (0.1)(0.96) \] \[ P(R) = 0.036 + 0.096 = 0.132 \]
05

Compute the Desired Probability

Now substitute the values into Bayes' theorem: \[ P(B \mid R) = \frac{P(R \mid B)P(B)}{P(R)} = \frac{0.036}{0.132} \] Simplifying, \[ P(B \mid R) = \frac{36}{132} = \frac{3}{11} \] So the probability is approximately \( 0.2727 \).

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Key Concepts

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

Conditional Probability
Conditional probability helps us find the likelihood of an event occurring based on the occurrence of another event. In our problem, we're interested in knowing the probability that Mary has a better hand, given that she decides to raise. This scenario involves calculating the probability of an event (Mary's hand being better) given another event (she raises) has happened already.

So, conditional probability is mathematically represented as:
  • \( P(A | B) \): the probability of event \( A \) happening given \( B \) has already occurred.
  • To compute this, we use the formula: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} \]
  • This approach helps update the probability of one event by considering the presence of another.
Without conditional probability, we wouldn't be able to accurately assess the situation based on new data, like in our poker hand problem.
Probability Theory
Probability theory is the study of randomness and uncertainty. It's a branch of mathematics dealing with the likelihood that different events will occur. Fundamental to probability theory are experiments, which are actions or processes that yield uncertain outcomes.

In our poker scenario:
  • We're examining the probability of Mary having a better hand than John.
  • This is an example of a probabilistic event based on the information we are provided, like Mary's decision to raise.
Probability theory gives us the tools to quantify and analyze the chances of events happening. This is crucial when making decisions based on incomplete information, such as determining the likelihood of Mary's hand being better based on her betting behavior.
Event Definition
In probability, an 'event' is any outcome or set of outcomes from a random process. In our example, specific events are being considered to address the problem.

Here's how event definition comes into play:
  • Event \( B \) is defined as "Mary has a better hand than John".
  • Event \( R \) is defined as "Mary raises".
Identifying and clearly defining events is crucial for setting up any probability problem efficiently. It allows us to use probability laws and theorems to compute the likelihood of complex outcomes, as we do in our poker hand problem. Without clear event definitions, applying probabilistic concepts becomes challenging.
Total Probability Theorem
The total probability theorem combines probabilities of different disjoint events to determine the likelihood of an overall event. It helps in calculating the overall probability of an event when it can occur from various independent scenarios.

In our example, we use the total probability theorem to find the probability that Mary raises, which involves combining the chances in two different cases:
  • Mary raises with a better hand.
  • Mary raises with a poorer hand.
Mathematically, this is expressed as:\[ P(R) = P(R \mid B)P(B) + P(R \mid B^c)P(B^c) \]This formula allows us to find \( P(R) \) by considering both scenarios, which is essential before applying Bayes' Theorem to find the conditional probability. Understanding this theorem is key for handling complex probability scenarios that depend on multiple conditions or events.

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Most popular questions from this chapter

(Johnsonbough \(^{8}\) ) A coin with probability \(p\) for heads is tossed \(n\) times. Let \(E\) be the event "a head is obtained on the first toss' and \(F_{k}\) the event 'exactly \(k\) heads are obtained." For which pairs \((n, k)\) are \(E\) and \(F_{k}\) independent?

Suppose that \(X\) and \(Y\) are continuous random variables with density functions \(f_{X}(x)\) and \(f_{Y}(y)\), respectively. Let \(f(x, y)\) denote the joint density function of \((X, Y)\). Show that $$ \int_{-\infty}^{\infty} f(x, y) d y=f_{X}(x) $$ and $$ \int_{-\infty}^{\infty} f(x, y) d x=f_{Y}(y) $$

Suppose that \(A\) and \(B\) are events such that \(P(A \mid B)=P(B \mid A)\) and \(P(A \cup B)=\) 1 and \(P(A \cap B)>0\). Prove that \(P(A)>1 / 2\).

You are given two urns each containing two biased coins. The coins in urn I come up heads with probability \(p_{1}\), and the coins in urn II come up heads with probability \(p_{2} \neq p_{1}\). You are given a choice of (a) choosing an urn at random and tossing the two coins in this urn or (b) choosing one coin from each urn and tossing these two coins. You win a prize if both coins turn up heads. Show that you are better off selecting choice (a).

A radioactive material emits \(\alpha\) -particles at a rate described by the density function \(f(t)=.1 e^{-.1 t}\) Find the probability that a particle is emitted in the first 10 seconds, given that (a) no particle is emitted in the first second. (b) no particle is emitted in the first 5 seconds. (c) a particle is emitted in the first 3 seconds. (d) a particle is emitted in the first 20 seconds.

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