/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 90 In California, about \(92 \%\) o... [FREE SOLUTION] | 91Ó°ÊÓ

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In California, about \(92 \%\) of teens who take the written driver's exam fail the first time they take it (www.teendrivingcourse.com). Suppose that Sam and Maria are randomly selected teenagers taking the test for the first time. a. What is the probability that they both pass the test? b. What is the probability that Sam OR Maria passes the test?

Short Answer

Expert verified
The probability that both Sam and Maria pass the test is \(0.0064\). The probability that either Sam or Maria passes the test is \(0.1536\).

Step by step solution

01

Identify failure and success probabilities

According to the exercise, the probability to fail the test is \(92\%\) or \(0.92\) in decimal. Therefore, the probability to pass the test is \(1 - 0.92 = 0.08\).
02

Calculate the probability that both pass the test

Since Sam and Maria are taking the test independently, we can calculate the probability of both passing by multiplying their individual probabilities of passing. So, the probability that both pass the test is \(0.08 * 0.08 = 0.0064\).
03

Calculate the probability that either one passes the test

The probability that either Sam or Maria passes can be determined by 1 minus the probability that they both fail the test. The probability of failing the test is \(0.92\). So the probability that they both fail is \(0.92 * 0.92 = 0.8464\). Therefore, the probability that at least one of them passes the test is \(1 - 0.8464 = 0.1536\).

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Key Concepts

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

Independent Events
Independent events are a fundamental concept in probability, particularly useful when evaluating multiple occurrences. For two or more events to be considered independent, the outcome of one event does not affect the outcome of the other. When examining situations such as Sam and Maria taking their driver's exams, each individual's performance on the test is not influenced by the other.
This independence allows us to analyze their probabilities separately. It’s essential to verify that the events you are considering are indeed independent before applying probability rules meant for independent events.
Think of independent events like flipping a fair coin; the result of each flip does not impact the next. Similarly, Sam passing the test doesn't increase or decrease Maria's chances of passing, and vice versa.
Complement Rule
In probability, the complement rule is a handy tool for solving problems. This rule states that the probability of an event happening is equal to one minus the probability of it not happening. If you know the likelihood of a specific outcome, calculating its complement is straightforward.
In our exercise, because we know the probability of failing the test, using the complement rule helps us find the probability of passing. So, if a failure probability is given as 0.92, the chance of passing, which is the complement, is easily found as 1 minus 0.92, or 0.08.
  • The complement rule simplifies calculations and is particularly useful when finding the probability of at least one event happening, as we've seen with Sam and Maria potentially passing.
Multiplication Rule for Independent Events
The multiplication rule of probability is used to find the likelihood of two or more independent events happening simultaneously. When events are independent, the probability that all of them occur is the product of their individual probabilities. This rule is extremely useful for simplifying calculations requiring multiple independent happenings.
Consider our example with Sam and Maria: Since each passing of the test is independent, we multiply their independent passing probabilities (0.08 each) to find out the probability of both passing, which is 0.08 * 0.08 = 0.0064.
  • This rule not only allows for more straightforward computation but also emphasizes the significance of independent probabilities in event chains.
  • Keep in mind that this rule only applies when events are truly independent and not when they influence each other in any way.

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Most popular questions from this chapter

One of the authors did a survey to determine the effect of students changing answers while taking a multiple-choice test on which there is only one correct answer for each question. Some students erase their initial choice and replace it with another. It turned out that \(61 \%\) of the changes were from incorrect answers to correct and that \(26 \%\) were from correct to incorrect. What percentage of changes were from incorrect to incorrect?

A certain professional basketball player typically makes \(80 \%\) of his basket attempts, which is considered to be good. Suppose you go to several games at which this player plays. Sometimes the player attempts only a few baskets, say, 10. Other times, he attempts about 60 . On which of those nights is the player most likely to have a "bad" night, in which he makes much fewer than \(80 \%\) of his baskets?

Independent Variables Use your general knowledge to label the following pairs of variables as independent or associated. Explain. a. For a sample of adults, gender and shoe size b. For a sample of football teams, win/loss record for the coin toss at the beginning of the game and number of cheerleaders for the team

Construct a two-way table with 80 men and 100 women in which both groups show an equal percentage of right-handedness.

The table shows the results of rolling a fair six-sided die. $$ \begin{array}{|c|c|c|c|} \hline \begin{array}{c} \text { Outcome } \\ \text { on Die } \end{array} & \mathbf{2 0} \text { Trials } & \mathbf{1 0 0} \text { Trials } & \mathbf{1 0 0 0} \text { Trials } \\ \hline 1 & 8 & 20 & 167 \\ \hline 2 & 4 & 23 & 167 \\ \hline 3 & 5 & 13 & 161 \\ \hline 4 & 1 & 13 & 166 \\ \hline 5 & 2 & 16 & 172 \\ \hline 6 & 0 & 15 & 167 \\ \hline \end{array} $$ Using the table, find the empirical probability of rolling a 1 for 20,100 , and 1000 trials. Report the theoretical probability of rolling a 1 with a fair six-sided die. Compare the empirical probabilities to the theoretical probability, and explain what they show.

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