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Chapter 2: Problem 71

If two events, \(A\) and \(B\), are such that \(P(A)=.5, P(B)=.3,\) and \(P(A \cap B)=.1\), find the following: a. \(P(A | B)\) b. \(P(B | A)\) c. \(P(A | A \cup B)\) d. \(P(A | A \cap B)\) e. \(P(A \cap B | A \cup B)\)

Short Answer

Expert verified
a. 0.333, b. 0.2, c. 0.714, d. 1, e. 0.143.

Step by step solution

01

Calculate P(A | B)

To find the probability of \(A\) given \(B\), use the formula for conditional probability: \[P(A | B) = \frac{P(A \cap B)}{P(B)}.\] Substituting the given values, we have: \[P(A | B) = \frac{0.1}{0.3} = \frac{1}{3} \approx 0.333.\]
02

Calculate P(B | A)

Similar to the previous step, use the formula for conditional probability: \[P(B | A) = \frac{P(A \cap B)}{P(A)}.\] Substituting the values, we get: \[P(B | A) = \frac{0.1}{0.5} = 0.2.\]
03

Calculate P(A | A 鈭� B)

First, find \(P(A \cup B)\) using the addition rule: \[P(A \cup B) = P(A) + P(B) - P(A \cap B).\] Substituting the values, \[P(A \cup B) = 0.5 + 0.3 - 0.1 = 0.7.\] Then apply the conditional probability formula: \[P(A | A \cup B) = \frac{P(A \cap (A \cup B))}{P(A \cup B)} = \frac{P(A)}{P(A \cup B)} = \frac{0.5}{0.7} \approx 0.714.\]
04

Calculate P(A | A 鈭� B)

Since \(A\) and \(A \cap B\) represent the same event when \(B\) is true, \(P(A | A \cap B) = 1\). It denotes the certainty of \(A\) given that both \(A\) and \(B\) happened.
05

Calculate P(A 鈭� B | A 鈭� B)

Use the conditional probability formula again: \[P(A \cap B | A \cup B) = \frac{P(A \cap B)}{P(A \cup B)}.\] Substituting the found values, \[P(A \cap B | A \cup B) = \frac{0.1}{0.7} \approx 0.143.\]

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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 events occurring. It gives us tools to make informed decisions and predictions based on chance. At its core, probability assigns a numerical value to how likely an event is to happen, ranging from 0 (impossible event) to 1 (certain event).

When solving problems using probability theory, we often examine different types of probabilities.
  • Simple probability: The chance of a single event occurring, such as flipping a head on a coin.

  • Compound probability: Deals with the probability of multiple events occurring together, such as drawing two hearts from a deck of cards in sequence.

  • Conditional probability: This calculates the likelihood of an event happening given that another event has already happened. This is crucial when events are dependent on one another.
Understanding these forms is vital for solving complex probability problems.
Joint Probability
Joint probability refers to the probability of two events happening at the same time. In mathematical terms, it looks like \(P(A \cap B)\).

When discussing probability, "joint" indicates that we are looking into the overlap of two events.
  • If the events A and B are independent, their joint probability is simply the product of their individual probabilities: \(P(A) \times P(B)\).

  • If the events are dependent, or if we have information about their overlap, it鈥檚 captured differently. We might know \(P(A \cap B)\), as given in a problem, which indicates this overlap directly.
Joint probability is crucial when trying to find overlapping areas between two probability sets and is pivotal in advanced statistical analyses.
Addition Rule in Probability
The Addition Rule in Probability helps us find the probability of either one event or another event occurring. It's particularly useful for non-mutually exclusive events, where both events could occur simultaneously.

The formula to calculate this is:
  • \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
This rule adds the probabilities of two events and subtracts the probability of their intersection (since it gets counted twice if both A and B are included).

The importance of the Addition Rule lies in its ability to correct for over-counting. By subtracting the intersection, it ensures the probability is accurate and not inflated by redundancies in counting.
Intersection and Union of Events
In probability, understanding the concepts of intersection and union is key.
  • Intersection (denoted as \(A \cap B\)) is the scenario where both events A and B occur. Conditional and joint probabilities often use this concept. It represents the common subset between two events.

  • Union (denoted as \(A \cup B\)), however, refers to any situation where either event A, event B, or both occur. The union is a broader scenario encompassing two or more events' likelihoods.
Being able to distinguish between these two concepts helps in understanding the range of possible outcomes within a probabilistic framework. It equips one with the skills to compute compound probabilities accurately and determine how often events might overlap or be exclusive.

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Most popular questions from this chapter

Suppose that two balanced dice are tossed repeatedly and the sum of the two uppermost faces is determined on each toss. What is the probability that we obtain a. a sum of 3 before we obtain a sum of 7? b. a sum of 4 before we obtain a sum of \(7 ?\)

Suppose two balanced coins are tossed and the upper faces are observed. a. List the sample points for this experiment. b. Assign a reasonable probability to each sample point. (Are the sample points equally likely?) c. Let \(A\) denote the event that exactly one head is observed and \(B\) the event that at least one head is observed. List the sample points in both \(A\) and \(B\). d. From your answer to part \((c),\) find \(P(A), P(B), P(A \cap B), P(A \cup B),\) and \(P(\bar{A} \cup B)\)

The symmetric difference between two events \(A\) and \(B\) is the set of all sample points that are in exactly one of the sets and is often denoted \(A \Delta B\). Note that \(A \Delta B=(A \cap B) \cup(\bar{A} \cap B)\). Prove that \(P(A \Delta B)=P(A)+P(B)-2 P(A \cap B)\)

The following game was played on a popular television show. The host showed a contestant three large curtains. Behind one of the curtains was a nice prize (maybe a new car) and behind the other two curtains were worthless prizes (duds). The contestant was asked to choose one curtain. If the curtains are identified by their prizes, they could be labeled \(G, D_{1},\) and \(D_{2}\) (Good Prize, Dud1, and Dud2). Thus, the sample space for the contestants choice is \(S=\left\\{G, D_{1}, D_{2}\right\\}\). a. If the contestant has no idea which curtains hide the various prizes and selects a curtain at random, assign reasonable probabilities to the simple events and calculate the probability that the contestant selects the curtain hiding the nice prize. b. Before showing the contestant what was behind the curtain initially chosen, the game show host would open one of the curtains and show the contestant one of the duds (he could always do this because he knew the curtain hiding the good prize). He then offered the contestant the option of changing from the curtain initially selected to the other remaining unopened curtain. Which strategy maximizes the contestant's probability of winning the good prize: stay with the initial choice or switch to the other curtain? In answering the following sequence of questions, you will discover that, perhaps surprisingly, this question can be answered by considering only the sample space above and using the probabilities that you assigned to answer part (a). i. If the contestant choses to stay with her initial choice, she wins the good prize if and only if she initially chose curtain \(G\). If she stays with her initial choice, what is the probability that she wins the good prize? ii. If the host shows her one of the duds and she switches to the other unopened curtain, what will be the result if she had initially selected \(G\) ? iii. Answer the question in part.(ii) if she had initially selected one of the duds. iv. If the contestant switches from her initial choice (as the result of being shown one of the duds), what is the probability that the contestant wins the good prize? v. Which strategy maximizes the contestant's probability of winning the good prize: stay with the initial choice or switch to the other curtain?

If \(A\) and \(B\) are equally likely events and we require that the probability of their intersection be at least.98, what is \(P(A) ?\)
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