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Chapter 3: Problem 4

Let \(P(B)>0 .\) Show that \(P(A \cap B)=P(A \mid B) P(B)\).

Short Answer

Expert verified
The equation \( P(A \cap B) = P(A \mid B) \cdot P(B) \) is derived from the definition of conditional probability given \( P(B) > 0 \).

Step by step solution

01

Understand Conditional Probability

First, let's recall the definition of conditional probability. The conditional probability of event A given event B is defined as \( P(A \mid B) = \frac{P(A \cap B)}{P(B)} \). This is the probability that event A occurs given that event B has already occurred.
02

Rearrange the Conditional Probability Formula

Given the formula for conditional probability \( P(A \mid B) = \frac{P(A \cap B)}{P(B)} \), we can rearrange it to solve for \( P(A \cap B) \). Multiplying both sides by \( P(B) \), we get:\[ P(A \cap B) = P(A \mid B) \cdot P(B) \].
03

Verify the Conditions

We were given the condition \( P(B) > 0 \). This condition ensures that dividing by \( P(B) \) in the conditional probability formula is valid (since division by zero is undefined). Since this condition holds true, our rearrangement in Step 2 is valid.

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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 branch of mathematics that deals with the likelihood of different outcomes. It's essential for understanding phenomena in various fields like finance, science, and everyday decision-making. Probability helps us quantify uncertainty and make informed predictions about future events.

One of the basic rules is that the probability of any event ranges from 0 to 1. This scale indicates the impossibility (0) to certainty (1) of the occurrence of an event.
  • A probability of 0 indicates that the event will not occur.
  • A probability of 1 means the event is certain to occur.
Probability theory uses this range to describe the chance of events happening. Understanding it is crucial to grasp more complex concepts like conditional probability, which considers the probability of an event occurring given the knowledge that another event has occurred.
Intersection of Events
The intersection of events relates to scenarios where we are interested in knowing the probability of two events occurring simultaneously. In probability notation, the intersection of event A and event B is represented as \(A \cap B\).
  • This concept is essential because it combines events to evaluate their joint probabilities.
  • For instance, if you want to find the probability of it raining and being a holiday on the same day, you'd calculate \(P(A \cap B)\).
For events A and B, the intersection \(A \cap B\) refers to situations where both events happen. In our original exercise, we used this concept to derive conditional probability.
Probability Formula
In probability theory, formulas help us calculate how likely an event is to happen based on known values. One such key formula is for conditional probability, which helps us understand the probability of an event given the occurrence of another event.
In the given exercise, we used the formula for conditional probability: \[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} \] This formula rearranges to show the relationship:\[ P(A \cap B) = P(A \mid B) \cdot P(B) \] Such rearrangements are essential in solving real-life problems where we need to find the likelihood of concurrent events.
The condition \(P(B) > 0 \) is crucial here because it ensures that division is possible (avoiding division by zero). Without such conditions, the operations would be mathematically invalid.

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Most popular questions from this chapter

Let \(A_{1}, A_{2}, \ldots, A_{n}\) be independent events. Show that the probability that none of the \(A_{1}, \ldots, A_{n}\) occur is less than or equal to \(\exp \left(-\sum_{i=1}^{n} P\left(A_{i}\right)\right)\).

Let \(P(C)>0\). Show that \(P(A \cup B \mid C)=P(A \mid C)+P(B \mid C)-P(A \cap B \mid\) C).

Suppose we model tossing a coin with two outcomes, \(H\) and \(T\), representing Heads and Tails. Let \(P(H)=P(T)=\frac{1}{2} .\) Suppose now we toss two such coins, so that the sample space of outcomes \(\Omega\) consists of four points: \(H H, H T, T H, T T\). We assume that the tosses are independent. a) Find the conditional probability that both coins show a head given that the first shows a head (answer: \(\frac{1}{2}\) ). b) Find the conditional probability that both coins show heads given that at least one of them is a head (answer: \(\frac{1}{3}\) ).

Donated blood is screened for AIDS. Suppose the test has \(99 \%\) accuracy, and that one in ten thousand people in your age group are HIV positive. The test has a \(5 \%\) false positive rating, as well. Suppose the test screens you as positive. What is the probability you have AIDS? Is it \(99 \% ?\) (Hint: \(99 \%\) refers to \(P\) (test positive|you have AIDS). You want to find \(P\) (you have AIDS|test is positive).

An urn contains \(r\) red balls and \(b\) blue balls. A ball is chosen at random from the urn, its color is noted, and it is returned together with \(d\) more balls of the same color. This is repeated indefinitely. What is the probability that a) The second ball drawn is blue? [Ans.: \(\left.\frac{b}{b+r}\right]\) b) The first ball drawn is blue given that the second ball drawn is blue?
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