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Sagot :
Let's tackle the problem step-by-step:
1. Understand the Given Data and Notations:
- [tex]\( M \)[/tex] represents the event that a vehicle has been driven many miles.
- [tex]\( B \)[/tex] represents the event that a vehicle is blue.
- The problem asks us to calculate [tex]\( P(B \mid M) \)[/tex].
2. Identify the Relevant Values from the Table:
- According to the table, the number of vehicles that have been driven many miles and are blue is [tex]\( 29 \)[/tex].
- The total number of vehicles that have been driven many miles is [tex]\( 71 \)[/tex].
3. Calculate the Conditional Probability [tex]\( P(B \mid M) \)[/tex]:
- The formula for conditional probability [tex]\( P(A \mid B) \)[/tex] is given by:
[tex]\[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} \][/tex]
- Here, [tex]\( A \)[/tex] is the event "the vehicle is blue" and [tex]\( B \)[/tex] is the event "the vehicle has been driven many miles."
So, we express the conditional probability [tex]\( P(B \mid M) \)[/tex] as:
[tex]\[ P(B \mid M) = \frac{\text{Number of vehicles that are blue and driven many miles}}{\text{Total number of vehicles driven many miles}} \][/tex]
4. Substitute the Known Values:
- The number of vehicles that are blue and driven many miles = [tex]\( 29 \)[/tex]
- The total number of vehicles driven many miles = [tex]\( 71 \)[/tex]
Thus,
[tex]\[ P(B \mid M) = \frac{29}{71} \][/tex]
5. Convert the Fraction to a Decimal:
- Performing the division, we get:
[tex]\[ \frac{29}{71} \approx 0.4084507042253521 \][/tex]
6. Interpret the Results:
- The calculated value [tex]\( P(B \mid M) = 0.4084507042253521 \)[/tex] implies that given a vehicle has been driven many miles, there is approximately a [tex]\( 0.408 \)[/tex] or [tex]\( 40.8\% \)[/tex] probability that the vehicle is blue.
Given this detailed explanation, the correct statement should be:
[tex]\( P(B \mid M) = 0.408 \)[/tex]; given that the vehicle has been driven many miles, there is approximately a 0.408 probability that the color is blue.
However, note that none of the original options exactly match our calculated conditional probability. Hence, there should be an error in the multiple-choice options provided. The correct interpretation of [tex]\( P(B \mid M) \)[/tex] based on our calculation is approximately [tex]\( 0.408 \)[/tex] or 40.8%.
1. Understand the Given Data and Notations:
- [tex]\( M \)[/tex] represents the event that a vehicle has been driven many miles.
- [tex]\( B \)[/tex] represents the event that a vehicle is blue.
- The problem asks us to calculate [tex]\( P(B \mid M) \)[/tex].
2. Identify the Relevant Values from the Table:
- According to the table, the number of vehicles that have been driven many miles and are blue is [tex]\( 29 \)[/tex].
- The total number of vehicles that have been driven many miles is [tex]\( 71 \)[/tex].
3. Calculate the Conditional Probability [tex]\( P(B \mid M) \)[/tex]:
- The formula for conditional probability [tex]\( P(A \mid B) \)[/tex] is given by:
[tex]\[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} \][/tex]
- Here, [tex]\( A \)[/tex] is the event "the vehicle is blue" and [tex]\( B \)[/tex] is the event "the vehicle has been driven many miles."
So, we express the conditional probability [tex]\( P(B \mid M) \)[/tex] as:
[tex]\[ P(B \mid M) = \frac{\text{Number of vehicles that are blue and driven many miles}}{\text{Total number of vehicles driven many miles}} \][/tex]
4. Substitute the Known Values:
- The number of vehicles that are blue and driven many miles = [tex]\( 29 \)[/tex]
- The total number of vehicles driven many miles = [tex]\( 71 \)[/tex]
Thus,
[tex]\[ P(B \mid M) = \frac{29}{71} \][/tex]
5. Convert the Fraction to a Decimal:
- Performing the division, we get:
[tex]\[ \frac{29}{71} \approx 0.4084507042253521 \][/tex]
6. Interpret the Results:
- The calculated value [tex]\( P(B \mid M) = 0.4084507042253521 \)[/tex] implies that given a vehicle has been driven many miles, there is approximately a [tex]\( 0.408 \)[/tex] or [tex]\( 40.8\% \)[/tex] probability that the vehicle is blue.
Given this detailed explanation, the correct statement should be:
[tex]\( P(B \mid M) = 0.408 \)[/tex]; given that the vehicle has been driven many miles, there is approximately a 0.408 probability that the color is blue.
However, note that none of the original options exactly match our calculated conditional probability. Hence, there should be an error in the multiple-choice options provided. The correct interpretation of [tex]\( P(B \mid M) \)[/tex] based on our calculation is approximately [tex]\( 0.408 \)[/tex] or 40.8%.
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