The probability that a U.S. resident has visited Canada is 0.18, the probalility that a U.S. resident has visited Mexico is 0.09, and the probability that a U.S. resident has visited both countries is 0.04. Consider the events "has visited Canada" and "has visited Mexico", as applied to a randomly-chosen U.S. resident. Are these two events independent? Are they disjoint? What can you say about these events? O They are independent but not disjoint o They are disjoint but not independent o They are both independent and disjoint o They are neither independent nor disjoint
The probability that a U.S. resident has visited Canada is 0.18, the probalility that a U.S. resident has visited Mexico is 0.09, and the probability that a U.S. resident has visited both countries is 0.04. Consider the events "has visited Canada" and "has visited Mexico", as applied to a randomly-chosen U.S. resident. Are these two events independent? Are they disjoint? What can you say about these events? O They are independent but not disjoint o They are disjoint but not independent o They are both independent and disjoint o They are neither independent nor disjoint
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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Transcribed Image Text:**Text for Educational Website**
**Probability Problem: Visiting Canada and Mexico**
The probability that a U.S. resident has visited Canada is 0.18. The probability that a U.S. resident has visited Mexico is 0.09. Furthermore, the probability that a U.S. resident has visited both countries is 0.04.
Consider the events "has visited Canada" and "has visited Mexico," applied to a randomly-chosen U.S. resident. Are these two events independent? Are they disjoint? What can you say about these events?
**Options:**
- They are independent but not disjoint.
- They are disjoint but not independent.
- They are both independent and disjoint.
- They are neither independent nor disjoint.
![**Question:**
The probability of team A beating team B is \(\frac{3}{5}\). What is the probability that team A will win two consecutive games from team B?
1. \(\frac{9}{25}\)
2. \(\frac{4}{25}\)
3. \(\frac{6}{25}\)
4. \(\frac{16}{25}\)
**Explanation:**
To find the probability of team A winning two consecutive games, you multiply the probability of winning one game by itself:
\[
\left(\frac{3}{5}\right) \times \left(\frac{3}{5}\right) = \frac{9}{25}
\]
Therefore, the correct answer is \(\frac{9}{25}\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F064aec2b-96bf-4be0-a1f2-689bafa9211a%2F886e17ab-8e76-41a1-a2ba-23e92b026bca%2Fe3yg29c_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question:**
The probability of team A beating team B is \(\frac{3}{5}\). What is the probability that team A will win two consecutive games from team B?
1. \(\frac{9}{25}\)
2. \(\frac{4}{25}\)
3. \(\frac{6}{25}\)
4. \(\frac{16}{25}\)
**Explanation:**
To find the probability of team A winning two consecutive games, you multiply the probability of winning one game by itself:
\[
\left(\frac{3}{5}\right) \times \left(\frac{3}{5}\right) = \frac{9}{25}
\]
Therefore, the correct answer is \(\frac{9}{25}\).
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