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91Ó°ÊÓ

Customers John runs a computer software store. Yesterday he counted 127 people who walked by his store, 58 of whom came into the store. Of the \(58,\) only 25 bought something in the store. (a) Estimate the probability that a person who walks by the store will enter the store. (b) Estimate the probability that a person who walks into the store will buy something. (c) Estimate the probability that a person who walks by the store will come in and buy something. (d) Estimate the probability that a person who comes into the store will buy nothing.

Short Answer

Expert verified
(a) ≈ 0.457, (b) ≈ 0.431, (c) ≈ 0.197, (d) ≈ 0.569.

Step by step solution

01

Identify Total and Relevant Groups for Part (a)

To estimate the probability that a person who walks by the store will enter, identify the total number of people who walked by as 127, and the number who entered the store as 58.
02

Calculate Probability for Part (a)

Use the formula for probability: \( P = \frac{\text{Number who entered the store}}{\text{Number who walked by}} \). Substitute the numbers: \( P = \frac{58}{127} \approx 0.457\). Thus, the probability is approximately 0.457.
03

Identify Numbers for Part (b)

For estimating the probability that a person who walks into the store will buy something, note that 25 people bought something out of the 58 who entered.
04

Calculate Probability for Part (b)

Use the formula: \( P = \frac{\text{Number who bought}}{\text{Number who entered}} \). Substitute the values: \( P = \frac{25}{58} \approx 0.431\). Therefore, the probability is approximately 0.431.
05

Identify Combined Events for Part (c)

To estimate the probability that a person who walks by will come in and buy something, note that there are 25 such people.
06

Calculate Combined Probability for Part (c)

Using the total number of people who walked by: \( P = \frac{25}{127} \approx 0.197\). Hence, the probability is approximately 0.197.
07

Identify Non-buying Group for Part (d)

To estimate the probability that a person who came into the store will buy nothing, note that 58 entered and 25 bought, so 33 did not buy.
08

Calculate Probability for Non-buyers in Part (d)

Use the formula: \( P = \frac{\text{Number who did not buy}}{\text{Number who entered}} \). Substitute: \( P = \frac{33}{58} \approx 0.569\). Thus, the probability is approximately 0.569.

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Key Concepts

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

Conditional Probability
Conditional probability is a critical concept in understanding likelihood in scenarios where some events are already known to have occurred. In this scenario, we're primarily interested in the probability of events given certain conditions are met.
If you want to find the probability that a person who enters John’s store will end up buying something, you are dealing with conditional probability. This is because your starting point is individuals who have already walked into the store; thus, it’s a subset of the larger group of people who walked by the store.

To calculate such probabilities, you use the conditionally limited sample size. In this case, you use the number of people who entered the store as your base. The probability equation would be:
  • \( P(\text{Buy | Enter}) = \frac{\text{Number who bought}}{\text{Total number who entered}} \)
This equation allows you to estimate the likelihood of the purchase event given that they have walked into the store. In this example, about 43.1% of individuals who enter end up purchasing something.
Basic Statistics
In basic statistics, probability is a fundamental aspect. It tells us how likely certain events are expected to occur. Probability values range from 0 to 1, where 0 indicates impossibility and 1 indicates certainty.
For calculating the probability of any event, you generally use the formula:
  • \( P = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \)

This formula helps estimate probabilities like those in John’s scenario, where you estimate that approximately 45.7% of people walking by his store enter it. Applying basic probability statistics also helps identify and solve problems related to purchase behavior, like determining how many walk-ins are likely to make a purchase.
Probability Estimation
Probability estimation is a way of using available data to predict future probabilities. This means taking past data, as we did with John's store visitors, to come up with a likelihood for future assessments.

For instance, estimating how many might enter and buy when 200 people walk by his store. By using previous data:
  • Walk-ins bought probability: \( P(\text{Buy and Enter}) = \frac{\text{Number bought}}{\text{Total walked by}} \)
  • In John’s case, this is \( \frac{25}{127} \approx 0.197 \)

This estimation shows approximately a 19.7% chance that someone will both enter and buy if they merely walk by the store. Probability estimation can help businesses like John’s predict buying trends and prepare sales strategies. It is valuable for planning resources and managing expectations regarding sales and customer interaction.

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Most popular questions from this chapter

Probability Estimate: Wiggle Your Ears Can you wiggle your ears? Use the students in your statistics class (or a group of friends) to estimate the percentage of people who can wiggle their ears. How can your result be thought of as an estimate for the probability that a person chosen at random can wiggle his or her ears? Comment: National statistics indicate that about \(13 \%\) of Americans can wiggle their ears (Source: Bernice Kanner, Are You Normal?, St. Martin's Press, New York).

Betting odds are usually stated against the event happening (against winning). The odds against event \(W\) are the ratio \(\frac{P(\text {not} W)}{P(W)}=\frac{P\left(W^{*}\right)}{P(W)}\) In horse racing, the betting odds are based on the probability that the horse does not win. (a) Show that if we are given the odds against an event \(W\) as \(a: b,\) the probability of not \(W\) is \(P\left(W^{\circ}\right)=\frac{a}{a+b} .\) Hint: Solve the cquation \(\frac{a}{b}=\frac{P(W)}{1-P\left(W^{\circ}\right)}\) for \(P\left(W^{c}\right)\) (b) In a recent Kentucky Derby, the betting odds for the favorite horse, Point Given, were 9 to \(5 .\) Use these odds to compute the probability that Point Given would lose the race. What is the probability that Point Given would win the race? (c) In the same race, the betting odds for the horse Monarchos were 6 to 1 . Use these odds to estimate the probability that Monarchos would lose the race. What is the probability that Monarchos would win the race? (d) Invisible Ink was a long shot, with betting odds of 30 to \(1 .\) Use these odds to estimate the probability that Invisible Ink would lose the race. What is the probability the horse would win the race? For further information on the Kentucky Derby, visit the web site of the Kentucky Derby.

Basic Computations: Rules of Probability Given \(P(A)=0.2, P(B)=0.5\) \(P(A | B)=0.3.\) (a) Compute \(P(A \text { and } B).\) (b) Compute \(P(A \text { or } B).\)

You roll two fair dice, a green one and a red one. (a) What is the probability of getting a sum of \(6 ?\) (b) What is the probability of getting a sum of \(4 ?\) (c) What is the probability of getting a sum of 6 or \(4 ?\) Are these outcomes mutually exclusive?

Compute \(C_{5,2}\).

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