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Chapter 4: Problem 8

Basic Computation: Rules of Probability Given \(P\left(A^{c}\right)=0.8, P(B)=0.3\) \(P(B | A)=0.2:\) (a) Compute \(P(A \text { and } B).\) (b) Compute \(P(A \text { or } B).\)

Short Answer

Expert verified
(a) \( P(A \text{ and } B) = 0.04 \), (b) \( P(A \text{ or } B) = 0.46 \).

Step by step solution

01

Find P(A)

To find the probability of event A, we need to find the complement of the probability given. We know that the probability of the complement of A, denoted as \( P(A^c) \), is 0.8. Since \( P(A) + P(A^c) = 1 \), we can calculate \( P(A) \) as follows:\[ P(A) = 1 - P(A^c) = 1 - 0.8 = 0.2 \].
02

Compute P(A and B)

We use the rule for conditional probability, where \( P(B | A) = \frac{P(A ext{ and } B)}{P(A)} \). We have \( P(B | A) = 0.2 \) and we just calculated \( P(A) = 0.2 \). Thus,\[ P(A ext{ and } B) = P(B | A) \times P(A) = 0.2 \times 0.2 = 0.04 \].
03

Compute P(A or B)

To find the probability of either event A or event B occurring, use the formula: \( P(A ext{ or } B) = P(A) + P(B) - P(A ext{ and } B) \). Now substitute the known values:\[ P(A ext{ or } B) = 0.2 + 0.3 - 0.04 = 0.46 \].

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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 understand how the probability of one event can change based on the occurrence of another event. Here, it shows the likelihood of event B occurring given that A has already happened. This is expressed as \( P(B | A) \). In general, the formula for conditional probability is:
  • \( P(B | A) = \frac{P(A \text{ and } B)}{P(A)} \)
The conditional probability \( P(B | A) \) is 0.2 in our exercise, meaning that once event A occurs, event B has a 20% chance of occurring as well. We used this concept to determine \( P(A \text{ and } B) \), the probability of both A and B happening together by rearranging the conditional probability formula:
  • \( P(A \text{ and } B) = P(B | A) \times P(A) = 0.2 \times 0.2 = 0.04 \)
Understanding conditional probability is key to finding how related events impact one another in terms of their probabilities.
Complementary Events
In probability theory, each event has a complementary event, which is everything that is not part of the original event. If you know the probability of an event's complement, you can easily find the probability of the event itself.
  • The relationship is simple: \( P(A) + P(A^c) = 1 \).
Given \( P(A^c) = 0.8 \) in the exercise, we calculated the probability of event A by subtracting the probability of its complement from 1:
  • \( P(A) = 1 - P(A^c) = 1 - 0.8 = 0.2 \)
Complementary events cover all possible outcomes of a situation. They are useful because they often make it easier to calculate rather than dealing with the main event directly, especially if you have the information on the complement probability readily available.
Addition Rule in Probability
The addition rule is a fundamental principle used to compute the probability of either one of two events occurring. It helps us find \( P(A \text{ or } B) \), which is the probability that either event A occurs, or event B occurs, or both occur together.
  • The formula is: \( P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B) \).
In the exercise, we had:
  • \( P(A) = 0.2 \)
  • \( P(B) = 0.3 \)
  • \( P(A \text{ and } B) = 0.04 \)
Using these values, the calculation was:
  • \( P(A \text{ or } B) = 0.2 + 0.3 - 0.04 = 0.46 \)
This rule is pivotal in understanding how probabilities of multiple events intertwine, ensuring you're not overcounting the probability where both events occur.

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Most popular questions from this chapter

Involve a standard deck of 52 playing cards. In such a deck of cards there are four suits of 13 cards each. The four suits are: hearts, diamonds, clubs, and spades. The 26 cards included in hearts and diamonds are red. The 26 cards included in clubs and spades are black. The 13 cards in each suit are: \(2,3,4,5,6,7,8,9,10,\) Jack, Queen, King, and Ace. This means there are four Aces, four Kings, four Queens, four \(10 \mathrm{s},\) etc., down to four \(2 \mathrm{s}\) in each deck. You draw two cards from a standard deck of 52 cards without replacing the first one before drawing the second. (a) Are the outcomes on the two cards independent? Why? (b) Find \(P(3 \text { on } 1 \text { st card and } 10 \text { on } 2 \text { nd })\) (c) Find \(P(10 \text { on } 1 \text { st card and } 3 \text { on } 2 \text { nd })\) (d) Find the probability of drawing a 10 and a 3 in either order.

You roll two fair dice, a green one and a red one. (a) Are the outcomes on the dice independent? (b) Find \(P(5 \text { on green die and } 3\) on red die ). (c) Find \(P(3 \text { on green die and } 5\) on red die ). (d) Find \(P[(5 \text { on green die and } 3 \text { on red die) or }(3 \text { on green die and } 5\text { on red die)]. }\)

Consider the following events for a college student selected at random: \(A=\) student is female \(B=\) student is majoring in business Translate each of the following phrases into symbols. (a) The probability the student is male or is majoring in business (b) The probability a female student is majoring in business (c) The probability a business major is female (d) The probability the student is female and is not majoring in business (e) The probability the student is female and is majoring in business

What is the law of large numbers? If you were using the relative frequency of an event to estimate the probability of the event, would it be better to use 100 trials or 500 trials? Explain.

(a) Draw a tree diagram to display all the possible head-tail sequences that can occur when you flip a coin three times. (b) How many sequences contain exactly two heads? (c) Probability Extension Assuming the sequences are all equally likely, what is the probability that you will get exactly two heads when you toss a coin three times?
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