91Ó°ÊÓ

A Web site uses ads to route visitors to one of four landing pages. The probabilities for each landing page are equal. Consider 20 independent visitors and let the random variables \(W\), \(X, Y,\) and \(Z\) denote the number of visitors routed to each page. Calculate the following: 5-44. A Web site uses ads to route visitors to one of four landing pages. The probabilities for each landing page are equal. Consider 20 independent visitors and let the random variables \(W\) \(X, Y,\) and \(Z\) denote the number of visitors routed to each page. Calculate the following: (a) \(P(W=5, X=5, Y=5, Z=5)\) (b) \(P(W=5, X=5, Y=5)\) (c) \(P(W=7, X=7, Y=6 \mid Z=3)\) (d) \(P(W=7, X=7, Y=3 \mid Z=3)\) (e) \(P(W \leq 2)\) (f) \(E(W)\) (g) \(P(W=5, X=5)\) (h) \(P(W=5 \mid X=5)\)

Step by step solution

01

Model Distribution

The problem involves routing 20 visitors equally among 4 landing pages. Hence, the random variables \( W, X, Y, Z \) follows a multinomial distribution \(\text{Multinomial}(n=20, p_1=p_2=p_3=p_4=0.25)\). Here, \( p = 0.25 \) indicates the probability that any visitor lands on one of the four pages.

02

Problem (a) Probability

For part (a), we need to calculate \( P(W=5, X=5, Y=5, Z=5) \). This is a multinomial probability given by:\[ P(W=5, X=5, Y=5, Z=5) = \frac{20!}{5!5!5!5!}\times(0.25)^5\times(0.25)^5\times(0.25)^5\times(0.25)^5 = \frac{20!}{5!5!5!5!}\times(0.25)^{20} \]Calculate using factorials and powers of 0.25.

03

Problem (b) Probability

For part (b), calculate \( P(W=5, X=5, Y=5) \), with no constraint on \( Z \). This is equivalent to counting all combinations such that \( W + X + Y = 15 \) where there's no direct need to account for \( Z \). Use the multinomial coefficient:\[ P(W=5, X=5, Y=5) = \frac{20!}{5!5!5!5!}\times(0.25)^{15}\times(1)^5 \] Effectively calculate it the same way but account for the sum condition.

04

Problem (c) Conditional Probability

For part (c), the conditional probability \( P(W=7, X=7, Y=6 \mid Z=3) \) involves evaluating only the event \( W+X+Y+Z = 20 \) given that \( Z=3 \).Since we have \( Z=3 \), we must have: \[ P(W=7, X=7, Y=6 \mid Z=3) = \frac{P(W=7, X=7, Y=6, Z=3)}{P(Z=3)} \] Substitute values: \[ P(W=7, X=7, Y=6, Z=3) = \frac{20!}{7!7!6!3!} \times (0.25)^{20} \]\[ P(Z=3) = \frac{20!}{3!17!} \times (0.25)^3 \times (0.75)^{17} \] Calculate these probabilities and divide them.

05

Problem (d) Conditional Probability

Part (d) \( P(W=7, X=7, Y=3 \mid Z=3) \) is calculated similarly, involving the same conditional framework but with a different setup: \[ P(W=7, X=7, Y=3 \mid Z=3) = \frac{P(W=7, X=7, Y=3, Z=3)}{P(Z=3)} \]Substitute values:\[ P(W=7, X=7, Y=3, Z=3) = \frac{20!}{7!7!3!3!} \times (0.25)^{20} \] Calculate and divide by \( P(Z=3) \) as in Step 4.

06

Problem (e) Cumulative Probability

Part (e) asks for \( P(W \leq 2) \). This requires summing probabilities:\[ P(W=0) + P(W=1) + P(W=2) \]Each \( P(W=k) \) for \( k = 0, 1, 2 \) involves binomial calculations given \( n = 20 \) and \( p = 0.25 \). Use:\[ P(W=k) = \binom{20}{k}(0.25)^k(0.75)^{20-k} \] Sum these values for the final probability.

07

Problem (f) Expectation

To find \( E(W) \), use the expected value of a multinomial distribution:\[ E(W) = n \times p = 20 \times 0.25 = 5 \] This is due to each page being targeted by 25% of the visitors.

08

Problem (g) Joint Probability

Part (g) \( P(W=5, X=5) \) asks for the joint probability that \( W = 5 \) and \( X = 5 \), with no constraint on \( Y \) or \( Z \), hence:\[ P(W=5, X=5) = \frac{20!}{5!5!(20-10)!} \times (0.25)^{10} \times (0.75)^{10} \] This reflects the choice of visitors being distributed without concern for \( Y \) and \( Z \)'s specifics while satisfying \( W+X = 10 \).

09

Problem (h) Conditional Probability

Calculate \( P(W=5 \mid X=5) \) using:\[ P(W=5 \mid X=5) = \frac{P(W=5, X=5)}{P(X=5)} \] Where \( P(W=5, X=5) \) was calculated in Step 8.\[ P(X=5) = \binom{20}{5}(0.25)^5(0.75)^{15} \] Compute and use this in the ratio for the conditional probability.

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Key Concepts

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

Probability Calculation

Understanding probability calculation is fundamental to solving any problem involving random events. In particular, when dealing with the multinomial distribution, the calculation becomes slightly more complex than the simple binomial distribution. The multinomial distribution describes the outcomes of independent events where each event can fall into one of several categories. Here:

• The probability of any visitor landing on one page is equal among all pages, meaning each probability is 0.25.
• We need to calculate probabilities of various combinations by applying the multinomial formula: \[ P(W=w, X=x, Y=y, Z=z) = \frac{n!}{w!x!y!z!} \times p^w \times p^x \times p^y \times p^z \]

For example, to find the probability of exactly 5 visitors landing on each page, you plug in these values for a straightforward calculation.
Combining factorials with exponentials of probabilities will give the likelihood of any specific configuration of events.

Independent Events

The concept of independent events is crucial in probability because it informs how probabilities are combined. Events are independent when the occurrence of one does not influence the likelihood of another.
In our case, each visitor landing on a page occurs independently of the others arriving at different or the same landing page. This independence ensures that probabilities for each visitor remain constant at 0.25 for landing on any particular page.
For example:

• If one visitor lands on Page W, it does not affect the probability for any subsequent visitor landing on Page W, X, Y, or Z.
• This property allows us to multiply probabilities for individual outcomes to find joint probabilities of multiple independent events.

This belief of independence underlies many ways we calculate probabilities, particularly when dealing with multinomial distributions like in this exercise.

Random Variables

In probability theory, a random variable is used to describe the outcomes of a random phenomenon. It assigns a numerical value to each event in a sample space.
In the problem at hand, we have four random variables: \( W, X, Y, \) and \( Z \). Each represents the number of visitors arriving at a particular landing page.

• These can take values from 0 to 20, as the total number of visitors is 20.
• They are governed by a probability distribution, in this case, the multinomial distribution.

You can think of these random variables as a way of organizing and quantifying the outcomes to make calculations easier. Computing probabilities, expectations, and variances become systematic using these variables, making them indispensable for precise probability analysis.

Conditional Probability

Conditional probability considers the likelihood of an event given that another event has already occurred, altering the sample space and affecting calculations. For example, calculating \( P(W=7, X=7, Y=6 \mid Z=3) \) involves assessing how the probability changes when it's known that \( Z=3 \) already occurred.
To compute conditional probabilities, you use:

• The formula: \( P(A \mid B) = \frac{P(A \cap B)}{P(B)} \)
• You must find the joint probability of all events occurring and then divide by the probability of the known event.

In our exercise, this computation allows us to understand not just the likelihood of specific visitor distributions across pages, but how other site metrics conditionally impact them. By focusing on known conditions, you gain insight into the effects of certain parameters on an entire probabilistic model.