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Chapter 11: Problem 81

In Exercises 81-85, write a probability problem involving the word "and" whose solution results in the probability fractions shown. \(\frac{1}{2} \cdot \frac{1}{2}\)

Short Answer

Expert verified
The probability problem could read: 'What is the probability of getting a head when flipping a fair coin twice?' The solution is \( \frac{1}{4} \).

Step by step solution

01

Understand the Problem

The aim is to write a probability problem using the word 'and', which when solved, results in the answer \( \frac{1}{2} \cdot \frac{1}{2} \). The word 'and' in probability signals that we need to multiply the likelihood of two independent events occurring.
02

Create the Problem Scenario

Let's consider a simple example of coin tossing. The problem could be: 'What is the probability of getting a head when flipping a balanced coin twice?'
03

Solve the Problem

The probability of getting a head in a single coin toss of a fair coin is \( \frac{1}{2} \). Since the two coin tosses are independent events, the probability of getting a head on both tosses is the product of the individual probabilities, which is \( \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \).

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Key Concepts

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

Independent Events
Understanding independent events is crucial in probability theory. Independent events are those whose outcomes do not affect each other. Put simply, the result of one event doesn't change the probability of the other event occurring. For example, when flipping a fair coin, getting heads on the first flip does not influence the result of the next flip. Each flip is an independent event, since coins have no memory and all flips are separate from each other.

Consider drawing a red card from a deck and then rolling a six on a dice. These two actions don't interfere with one another; the deck doesn't 'know' what happened with the dice and vice versa. Recognizing independent events is essential for accurately calculating probabilities, especially when multiple events are involved.
Multiplication Rule in Probability
The multiplication rule is a fundamental concept in determining the likelihood of multiple independent events occurring together. It states that the probability of the conjunction of two independent events is equal to the product of their individual probabilities.

If 'Event A' has a probability of occurring of \( P(A) \) and 'Event B' has a probability of occurring of \( P(B) \) and both events are independent, the probability that both 'Event A' and 'Event B' will occur (denoted as \( P(A \text{ and } B) \) or \( P(A \cap B) \) ) is \( P(A) \times P(B) \). This is an important tool in solving complex probability problems because it simplifies the analysis of multiple actions taking place.
Coin Toss Probability
Coin toss probability is a classic example used to teach basic probability. A fair coin has two sides, heads and tails, and if the coin is fair, the chance of landing on heads is the same as the chance of landing on tails, which is \( \frac{1}{2} \).

In our example, we look at the probability of flipping heads twice in a row. Since we're dealing with a fair coin and tossing it twice are two independent events, we use the multiplication rule. The chance of flipping heads on the first toss is \( \frac{1}{2} \) and likewise \( \frac{1}{2} \) for the second toss. Therefore, the combined probability is \( \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \). This simple yet powerful example shows how the multiplication rule applies to real-world situations and helps in understanding more complex probability scenarios.

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Most popular questions from this chapter

I found the probability of getting rain at least once in ten days by calculating the probability that none of the days have rain and subtracting this probability from \(1 .\)

The tables in Exercises 3-4 show claims and their probabilities for an insurance company. a. Calculate the expected value and describe what this means in practical terms. b. How much should the company charge as an average premium so that it breaks even on its claim costs? c. How much should the company charge to make a profit of \(\$ 50\) per policy? PROBABILITIES FOR MEDICAL INSURANCE CLAIMS $$ \begin{array}{|c|c|} \hline \begin{array}{c} \text { Amount of Claim (to the } \\ \text { nearest } \$ 20,000) \end{array} & \text { Probability } \\ \hline \$ 0 & 0.70 \\ \hline \$ 20,000 & 0.20 \\ \hline \$ 40,000 & 0.06 \\ \hline \$ 60,000 & 0.02 \\ \hline \$ 80,000 & 0.01 \\ \hline \$ 100,000 & 0.01 \\ \hline \end{array} $$

Here's my dilemma: I can accept a \(\$ 1000\) bill or play a dice game ten times. For each roll of the single die, \- I win \(\$ 500\) for rolling 1 or 2 . \- I win \(\$ 200\) for rolling 3 . \- I lose \(\$ 300\) for rolling 4,5 , or 6 . Based on expected value, I should accept the \(\$ 1000\) bill.

It is estimated that there are 27 deaths for every 10 million people who use airplanes. A company that sells flight insurance provides \(\$ 100,000\) in case of death in a plane crash. A policy can be purchased for \(\$ 1\). Calculate the expected value and thereby determine how much the insurance company can make over the long run for each policy that it sells.

A 25 -year-old can purchase a one-year life insurance policy for \(\$ 10,000\) at a cost of \(\$ 100\). Past history indicates that the probability of a person dying at age 25 is \(0.002\). Determine the company's expected gain per policy.
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