Asked by Alex
Use the table to answer the question.
Arrival Time
Transport Type On Time Late TOTAL
Car 230 20 250
Bicycle 70 130 200
TOTAL 300 150 450
Let event A
represent a late arrival and event B
represent transportation by car. Which of the following shows the correct calculation of P(A)⋅P(B)
?
(1 point)
Responses
150250+20250
Start Fraction 150 over 250 End Fraction plus Start Fraction 20 over 250 End Fraction
150250⋅20250
Start Fraction 150 over 250 End Fraction times Start Fraction 20 over 250 End Fraction
150450+250450
Start Fraction 150 over 450 End Fraction plus Start Fraction 250 over 450 End Fraction
150450⋅250450
Arrival Time
Transport Type On Time Late TOTAL
Car 230 20 250
Bicycle 70 130 200
TOTAL 300 150 450
Let event A
represent a late arrival and event B
represent transportation by car. Which of the following shows the correct calculation of P(A)⋅P(B)
?
(1 point)
Responses
150250+20250
Start Fraction 150 over 250 End Fraction plus Start Fraction 20 over 250 End Fraction
150250⋅20250
Start Fraction 150 over 250 End Fraction times Start Fraction 20 over 250 End Fraction
150450+250450
Start Fraction 150 over 450 End Fraction plus Start Fraction 250 over 450 End Fraction
150450⋅250450
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To calculate \(P(A) \cdot P(B)\), we need to start by determining the probabilities of events \(A\) and \(B\).
1. **Calculating \(P(A)\)** (probability of a late arrival):
- There are 150 late arrivals out of a total of 450 arrivals.
- Therefore, \(P(A) = \frac{150}{450}\).
2. **Calculating \(P(B)\)** (probability of transportation by car):
- There are 250 total car arrivals out of a total of 450 arrivals.
- Therefore, \(P(B) = \frac{250}{450}\).
3. **Calculating \(P(A) \cdot P(B)\)**:
- Now, we multiply these probabilities:
\[
P(A) \cdot P(B) = \left(\frac{150}{450}\right) \cdot \left(\frac{250}{450}\right)
\]
This matches the option:
- \( \frac{150}{450} \cdot \frac{250}{450} \)
Thus, the correct response is:
**150450⋅250450**
\( \text{Start Fraction 150 over 450 End Fraction times Start Fraction 250 over 450 End Fraction} \)
1. **Calculating \(P(A)\)** (probability of a late arrival):
- There are 150 late arrivals out of a total of 450 arrivals.
- Therefore, \(P(A) = \frac{150}{450}\).
2. **Calculating \(P(B)\)** (probability of transportation by car):
- There are 250 total car arrivals out of a total of 450 arrivals.
- Therefore, \(P(B) = \frac{250}{450}\).
3. **Calculating \(P(A) \cdot P(B)\)**:
- Now, we multiply these probabilities:
\[
P(A) \cdot P(B) = \left(\frac{150}{450}\right) \cdot \left(\frac{250}{450}\right)
\]
This matches the option:
- \( \frac{150}{450} \cdot \frac{250}{450} \)
Thus, the correct response is:
**150450⋅250450**
\( \text{Start Fraction 150 over 450 End Fraction times Start Fraction 250 over 450 End Fraction} \)
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