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91Ó°ÊÓ

Facebook versus YouTube A recent survey suggests that \(85 \%\) of college students have posted a profile on Facebook, \(73 \%\) use YouTube regularly, and \(66 \%\) do both. Suppose we select a college student at random and learn that the student has a profile on Facebook. Find the probability that the student uses YouTube regularly. Show your work.

Short Answer

Expert verified
The probability is approximately 77.65%.

Step by step solution

01

Define the Sets

Let's define two sets: Let \( F \) be the event that a student has a profile on Facebook, and \( Y \) be the event that the student uses YouTube regularly. From the problem, we have \( P(F) = 0.85 \), \( P(Y) = 0.73 \), and \( P(F \cap Y) = 0.66 \).
02

Apply Conditional Probability Formula

We want to find the probability that a student uses YouTube regularly given that they have a Facebook profile, which is represented as \( P(Y|F) \). According to the formula for conditional probability, \( P(Y|F) = \frac{P(Y \cap F)}{P(F)} \).
03

Substitute Known Values

Now we substitute the known values into the conditional probability formula: \( P(Y|F) = \frac{P(Y \cap F)}{P(F)} = \frac{0.66}{0.85} \).
04

Calculate the Probability

Perform the division \( \frac{0.66}{0.85} \) to find the probability. This calculation gives approximately \( 0.7765 \).
05

Interpret the Result

Thus, the probability that a college student uses YouTube regularly given that they have a Facebook profile is approximately \( 77.65\% \).

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Key Concepts

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

Venn diagrams
Venn diagrams are a fantastic way to visually represent relationships between different sets. In this context, a Venn diagram helps us understand the overlap between students who have a profile on Facebook and those who use YouTube regularly. Imagine two circles: one labeled Facebook (\( F \)) and the other YouTube (\( Y \)). The circle labeled \( F \) includes all students with a Facebook profile, and \( Y \) includes those who use YouTube regularly.

Where these circles overlap, we find students who both have a Facebook profile and use YouTube (\( F \cap Y \)). The Venn diagram not only shows individual probabilities like \( P(F) = 0.85 \) (85% have Facebook) and \( P(Y) = 0.73 \) (73% use YouTube), but importantly, it highlights the intersection where \( P(F \cap Y) = 0.66 \) (66% do both).

This overlap area is critical in understanding how the two groups relate, especially when calculating conditional probabilities. By seeing these overlaps visually, the complex relationships between sets become much easier to comprehend!
mutually exclusive events
Mutually exclusive events are events that cannot happen at the same time. For example, if you roll a die, getting a 3 and a 5 on the same roll are mutually exclusive events—the occurrence of one excludes the occurrence of the other.

In the context of students using Facebook and YouTube, these events are **not** mutually exclusive. This is because a student can certainly have a Facebook profile and use YouTube regularly at the same time. The problem explicitly tells us that 66% of students do both, indicating overlap in this case.

Understanding whether events are mutually exclusive is vital in determining how to approach probability calculations. Knowledge that two events are not mutually exclusive allows us to consider their intersection, crucial for calculating conditional probabilities.
probability calculations
Probability calculations often start with basic probability definitions but can extend to more complex concepts when events overlap or depend on each other. The conditional probability is one such concept and is represented by \( P(A|B) \), the probability of event A happening given that event B has occurred.

In the given survey, we wanted to find \( P(Y|F) \), the probability a student uses YouTube given they have a Facebook profile. Using the formula for conditional probability, \( P(Y|F) = \frac{P(Y \cap F)}{P(F)} \), we substitute \( P(Y \cap F) = 0.66 \) and \( P(F) = 0.85 \).

By performing the division \( \frac{0.66}{0.85} \), we arrive at approximately 0.7765, which tells us that about 77.65% of Facebook profile holders use YouTube regularly. Probability calculations like these allow us to make informed conclusions about relationships within data.

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