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A Web site randomly selects among 10 products to discount each day. The color printer of interest to you is discounted today. a. What is the expected number of days until this product is again discounted? b. What is the probability that this product is first discounted again exactly 10 days from now? c. If the product is not discounted for the next five days, what is the probability that it is first discounted again 15 days from now? d. What is the probability that this product is first discounted again within three or fewer days?

Short Answer

Expert verified
a) 10 days, b) 0.0387, c) 0.0387, d) 0.271

Step by step solution

01

Understanding the Problem

We need to solve a problem related to waiting times and probabilities associated with discounts for a specific product. The discount selection among products is random, with a color printer (the product of interest) being one of these choices once among ten. Thus, each day presents an independent 1/10 chance the printer gets a discount.
02

Expected Days until Discount

The expected number of days until the product is discounted follows a geometric distribution where the probability of success (getting a discount) is 0.1 (1/10). The expected number of days for a discount is given by the reciprocal of the probability: \[ E(X) = \frac{1}{p} = \frac{1}{0.1} = 10 \text{ days} \]
03

Probability of Discount in Exactly 10 Days

The probability that the product is first discounted again exactly after 10 days follows a geometric distribution with a success probability of 0.1. The formula for this is: \[ P(X = 10) = (1-p)^{n-1} \cdot p = 0.9^9 \cdot 0.1 \approx 0.0387 \]
04

Probability of Discount After 15 Days, Given 5 Days Passed

Given that the product hasn't been discounted for 5 days, we want the probability it first gets discounted on the 15th day—that is not discounted for the next 9 days (from day 6 to day 14) and then discounted. This conditional probability is still based on the memorylessness property of the geometric distribution: \[ P(X = 15 | X > 5) = (1-p)^{15-5-1} \cdot p = 0.9^9 \cdot 0.1 \approx 0.0387 \]
05

Probability of Being Discounted Within Three Days

To find the probability of the product being discounted within three days, sum up the probabilities for 1, 2, and 3 days: \[ P(X \leq 3) = P(X=1) + P(X=2) + P(X=3) = 0.1 + 0.9\times0.1 + 0.9^2\times0.1 \] Calculating gives: \[ P(X \leq 3) = 0.1 + 0.09 + 0.081 = 0.271 \]

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Key Concepts

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

Probability of Discount
The concept of probability in the context of discounts can be seen as a way of measuring the chance of an event happening. For the color printer discount scenario, this probability is straightforward. With the web site randomly selecting one of the ten products to discount each day, the probability of the printer being chosen for a discount on any given day is \( \frac{1}{10} \) or 0.1.
  • This probability is constant and does not depend on previous outcomes, which is a key aspect of random selection.
  • It is this constant probability that helps in calculating other associated probabilities using the geometric distribution.
Understanding this basic probability allows us to further explore its implications in expected days and the chance of future discounts.
Expected Value
Expected value is a fundamental concept in probability that gives us a measure of the center of a probability distribution. In simpler terms, it tells us what average outcome one can expect over a long run of experiments. For a geometric distribution, which models the number of trials until the first success, the expected value is given by the formula: \( E(X) = \frac{1}{p} \).

Applying Expected Value

If the probability of the color printer being discounted each day is 0.1, the expected number of days until this event occurs again is calculated as \( \frac{1}{0.1} = 10 \) days.
  • This means that, on average, you would expect to wait ten days between discounts for the same product.
  • It’s important to remember that this "average" expectation doesn't guarantee that every discount will occur exactly after ten days, but rather it's a statistical average over many cycles.
Memorylessness Property
The memorylessness property is a special feature of the geometric distribution, which states that the probability of an event occurring is the same regardless of how long it has already been elapsed without the event happening. Simply put, waiting for a further event is independent of past events.

Understanding the Impact of Memorylessness

In the context of the web site's discount practices, suppose the color printer has not been discounted for the past five days. Imagine now you want to find the probability it is discounted exactly on the 15th day from now. Memorylessness implies that it doesn't matter what happened in the past five days; the probability of being discounted starting from any prior point remains the same.
  • This is calculated just like calculating a new geometric probability: \( P(X = 15 | X > 5) = (1-p)^{15-5-1} \cdot p = 0.9^9 \cdot 0.1 \).
  • This feature simplifies calculations and predictions about future discount events when using the geometric distribution model.
Memorylessness can be counterintuitive at first but is a powerful tool when modeling scenarios like repeated games or random selections.

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