Find the indicated probability. 80% of U.S. drivers over the age of 30 have had a speeding ticket and 55% of tho have also been in an accident while behind the wheel. What is the probability that a randomly selected U.S. driver over the age of 30 ha had a speeding ticket and also been in an accident while behind the wheel? O 0.64 O 0.6875 O 0.3025 O 0.25 O 0.44

MATLAB: An Introduction with Applications
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Author:Amos Gilat
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19)
**Find the indicated probability.**

80% of U.S. drivers over the age of 30 have had a speeding ticket and 55% of those have also been in an accident while behind the wheel.

What is the probability that a randomly selected U.S. driver over the age of 30 has had a speeding ticket and also been in an accident while behind the wheel?

- ○ 0.64
- ○ 0.6875
- ○ 0.3025
- ○ 0.25
- ○ 0.44  

**Solution Explanation:**

To find the probability, we multiply the probability of each independent event. 

Let \( A \) be the event "having a speeding ticket," and \( B \) be the event "been in an accident."

Given,
- \( P(A) = 0.80 \)
- \( P(B|A) = 0.55 \)

We want to find \( P(A \cap B) \), which is \( P(A) \times P(B|A) \).

Calculating:
\[ 
P(A \cap B) = 0.80 \times 0.55 = 0.44 
\]

Thus, the probability that a randomly selected U.S. driver over the age of 30 has had a speeding ticket and also been in an accident while behind the wheel is **0.44**.
Transcribed Image Text:**Find the indicated probability.** 80% of U.S. drivers over the age of 30 have had a speeding ticket and 55% of those have also been in an accident while behind the wheel. What is the probability that a randomly selected U.S. driver over the age of 30 has had a speeding ticket and also been in an accident while behind the wheel? - ○ 0.64 - ○ 0.6875 - ○ 0.3025 - ○ 0.25 - ○ 0.44 **Solution Explanation:** To find the probability, we multiply the probability of each independent event. Let \( A \) be the event "having a speeding ticket," and \( B \) be the event "been in an accident." Given, - \( P(A) = 0.80 \) - \( P(B|A) = 0.55 \) We want to find \( P(A \cap B) \), which is \( P(A) \times P(B|A) \). Calculating: \[ P(A \cap B) = 0.80 \times 0.55 = 0.44 \] Thus, the probability that a randomly selected U.S. driver over the age of 30 has had a speeding ticket and also been in an accident while behind the wheel is **0.44**.
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