Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Suppose that at a particular supermarket the probability of waiting 5 minutes or longer for checkout at the cashier's counter is . \(2 .\) On a given day, a man and his wife decide to shop individually at the market, each checking out at different cashier counters. They both reach cashier counters at the same time. a. What is the probability that the man will wait less than 5 minutes for checkout? b. What is probability that both the man and his wife will be checked out in less than 5 minutes? (Assume that the checkout times for the two are independent events.) c. What is the probability that one or the other or both will wait 5 minutes or longer?

Short Answer

Expert verified
Answer: The probability that one or the other or both will wait 5 minutes or longer at the cashier's counter is 0.36.

Step by step solution

01

Complementary Probability for the man

The probability of waiting 5 minutes or longer for checkout is 0.2. To find the probability that the man will wait less than 5 minutes, we need to find the complement of this event, which is the probability of not waiting 5 minutes or longer. To find the complementary probability, we subtract the given probability from 1: Man waiting less than 5 minutes = 1 - Probability of waiting 5 minutes or more = 1 - 0.2 = 0.8
02

Probability of both waiting less than 5 minutes

Both the man and his wife's checkout times are independent events. Therefore, to find the probability of both of them waiting less than 5 minutes, we can multiply their individual probabilities: P(Both waiting less than 5 minutes) = P(Man waiting less than 5 minutes) × P(Wife waiting less than 5 minutes) = 0.8 × 0.8 = 0.64
03

Probability of one or the other or both waiting 5 minutes or longer

To find the probability of one or the other or both waiting 5 minutes or longer, we can first find the probability of both waiting less than 5 minutes, which we already found in Step 2. Then, we use the complementary probability to find the answer to this part: P(One or the other or both waiting 5 minutes or longer) = 1 - P(Both waiting less than 5 minutes) = 1 - 0.64 = 0.36 To summarize the results: a. The probability that the man will wait less than 5 minutes is 0.8 b. The probability that both the man and his wife will be checked out in less than 5 minutes is 0.64 c. The probability that one or the other or both will wait 5 minutes or longer is 0.36

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Complementary Probability
In probability, the concept of complementary events is a powerful tool that simplifies calculations. The complement of an event is everything that is not the event itself. Hence, to find the complementary probability, you subtract the probability of the event from 1. For example, in our supermarket scenario, the probability that the man will wait five minutes or longer is 0.2. Therefore, the probability of the complement event, which is the man waiting less than five minutes, can be calculated by subtracting 0.2 from 1, giving us 0.8. This means there is an 80% chance he will be checked out in under five minutes.
Some key points about complementary probabilities:
  • The sum of the probabilities of an event and its complement is always equal to 1.
  • Complementary probability is useful in scenarios where it is easier to calculate the probability of an event not occurring.
Independent Events
Understanding independent events is crucial in probability. Two events are considered independent if the outcome of one event does not influence the outcome of the other. This property is pivotal when dealing with multiple random events.In the example of the man and his wife shopping separately, their checkout times are independent events. This means that the probability of one person being checked out in less than five minutes is independent of the other person's checkout time.
When events are independent:
  • The occurrence of one event provides no information about the occurrence of the other event.
  • The probability of both events occurring is the product of their individual probabilities.
To illustrate this, the probability that both the man and his wife are checked out in less than five minutes is calculated by multiplying their individual probabilities: \( P(\text{Both less than 5 minutes}) = P(\text{Man less than 5}) \times P(\text{Wife less than 5}) = 0.8 \times 0.8 = 0.64 \) This result, 0.64, shows that there is a 64% chance both individuals will finalize checkout in under five minutes.
Multiplication Rule for Probabilities
The multiplication rule for probabilities is essential when working with scenarios involving several events. This rule simplifies the calculation of the probability of multiple independent events occurring together. The multiplication rule states that for independent events \( A \) and \( B \), the probability of both events occurring, \( P(A \text{ and } B) \), is the product of their separate probabilities: \[ P(A \cap B) = P(A) \times P(B) \]Applying this rule to our problem, where both the man and his wife check out at separate counters, we calculate the joint probability of both waiting less than five minutes by multiplying each individual's probability of waiting less than five minutes, which is 0.8 for both. Thus, \( P(\text{Both less than 5 minutes}) = 0.8 \times 0.8 = 0.64 \)Key takeaways from the multiplication rule for probabilities include:
  • The multiplication rule is only applicable to independent events.
  • It helps in calculating the simultaneous occurrence of two or more events.
  • Real-world examples often involve independent events, such as flipping a coin or rolling a die.
Overall, understanding the multiplication rule aids in solving complex problems involving multiple steps or layers of probability.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A food company plans to conduct an experiment to compare its brand of tea with that of two competitors. A single person is hired to taste and rank each of three brands of tea, which are unmarked except for identifying symbols \(A, B\), and \(C\). a. Define the experiment. b. List the simple events in \(S\). c. If the taster has no ability to distinguish a difference in taste among teas, what is the probability that the taster will rank tea type \(A\) as the most desirable? As the least desirable?

Medical case histories indicate that different illnesses may produce identical symptoms. Suppose a particular set of symptoms, which we will denote as event \(H,\) occurs only when any one of three illnesses \(-A, B\), or \(C\) - occurs. (For the sake of simplicity, we will assume that illnesses \(A\), \(B\), and \(C\) are mutually exclusive.) Studies show these probabilities of getting the three illnesses: $$\begin{array}{l}P(A)=.01 \\\P(B)=.005 \\\P(C)=.02\end{array}$$ The probabilities of developing the symptoms \(H\), given a specific illness, are $$\begin{array}{l}P(H \mid A)=.90 \\\P(H \mid B)=.95 \\\P(H \mid C)=.75\end{array}$$ Assuming that an ill person shows the symptoms \(H\), what is the probability that the person has illness \(A\) ?

Two tennis professionals, \(A\) and \(B\), are scheduled to play a match; the winner is the first player to win three sets in a total that cannot exceed five sets. The event that \(A\) wins any one set is independent of the event that \(A\) wins any other, and the probability that \(A\) wins any one set is equal to .6. Let \(x\) equal the total number of sets in the match; that is, \(x=3,4,\) or \(5 .\) Find \(p(x)\).

Player \(A\) has entered a golf tournament but it is not certain whether player \(B\) will enter. Player \(A\) has probability \(1 / 6\) of winning the tournament if player \(B\) enters and probability \(3 / 4\) of winning if player \(B\) does not enter the tournament. If the probability that player \(B\) enters is \(1 / 3,\) find the probability that player \(A\) wins the tournament.

Under the "no pass, no play" rule for athletes, an athlete who fails a course is disqualified from participating in sports activities during the next grading period. Suppose the probability that an athlete who has not previously been disqualified will be disqualified is .15 and the probability that an athlete who has been disqualified will be disqualified again in the next time period is \(.5 .\) If \(30 \%\) of the athletes have been disqualified before, what is the unconditional probability that an athlete will be disqualified during the next grading period?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free