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Sagot :
Let's solve this step-by-step.
### Given Data:
We are provided with percentages regarding the use of seat-belts and the percentage of drivers stopped for moving violations in two groups:
- Group I (using seat-belts):
- Percentage of drivers in this group: [tex]\( 61\% \)[/tex]
- Percentage of drivers in this group stopped for a moving violation: [tex]\( 0.4\% \)[/tex]
- Group II (not using seat-belts):
- Percentage of drivers in this group: [tex]\( 39\% \)[/tex]
- Percentage of drivers in this group stopped for a moving violation: [tex]\( 0.3\% \)[/tex]
### Step-by-Step Solution:
#### Part (a): Probability that a stopped driver will have a seat-belt on.
1. Convert percentages to decimals:
- [tex]\( P(\text{Seat-belt on}) = 0.61 \)[/tex]
- [tex]\( P(\text{No seat-belt}) = 0.39 \)[/tex]
- [tex]\( P(\text{Stopped with seat-belt}) = 0.004 \)[/tex]
- [tex]\( P(\text{Stopped without seat-belt}) = 0.003 \)[/tex]
2. Calculate the total probability of being stopped:
[tex]\[ P(\text{Stopped}) = P(\text{Seat-belt on}) \times P(\text{Stopped with seat-belt}) + P(\text{No seat-belt}) \times P(\text{Stopped without seat-belt}) \][/tex]
[tex]\[ P(\text{Stopped}) = 0.61 \times 0.004 + 0.39 \times 0.003 = 0.00244 + 0.00117 = 0.00361 \][/tex]
3. Calculate the probability of having a seat-belt on when stopped:
[tex]\[ P(\text{Seat-belt on} \mid \text{Stopped}) = \frac{P(\text{Seat-belt on}) \times P(\text{Stopped with seat-belt})}{P(\text{Stopped})} \][/tex]
[tex]\[ P(\text{Seat-belt on} \mid \text{Stopped}) = \frac{0.61 \times 0.004}{0.00361} = \frac{0.00244}{0.00361} \approx 0.676 \][/tex]
Therefore, the probability that a stopped driver will have a seat-belt on is approximately [tex]\( 0.676 \)[/tex].
#### Part (b): Probability that a stopped driver will not have a seat-belt on.
1. Calculate the probability of not having a seat-belt on when stopped:
[tex]\[ P(\text{No seat-belt} \mid \text{Stopped}) = \frac{P(\text{No seat-belt}) \times P(\text{Stopped without seat-belt})}{P(\text{Stopped})} \][/tex]
[tex]\[ P(\text{No seat-belt} \mid \text{Stopped}) = \frac{0.39 \times 0.003}{0.00361} = \frac{0.00117}{0.00361} \approx 0.324 \][/tex]
Therefore, the probability that a stopped driver will not have a seat-belt on is approximately [tex]\( 0.324 \)[/tex].
### Summary:
- (a) The probability that a stopped driver will have a seat-belt on is approximately [tex]\( 0.676 \)[/tex].
- (b) The probability that a stopped driver will not have a seat-belt on is approximately [tex]\( 0.324 \)[/tex].
### Given Data:
We are provided with percentages regarding the use of seat-belts and the percentage of drivers stopped for moving violations in two groups:
- Group I (using seat-belts):
- Percentage of drivers in this group: [tex]\( 61\% \)[/tex]
- Percentage of drivers in this group stopped for a moving violation: [tex]\( 0.4\% \)[/tex]
- Group II (not using seat-belts):
- Percentage of drivers in this group: [tex]\( 39\% \)[/tex]
- Percentage of drivers in this group stopped for a moving violation: [tex]\( 0.3\% \)[/tex]
### Step-by-Step Solution:
#### Part (a): Probability that a stopped driver will have a seat-belt on.
1. Convert percentages to decimals:
- [tex]\( P(\text{Seat-belt on}) = 0.61 \)[/tex]
- [tex]\( P(\text{No seat-belt}) = 0.39 \)[/tex]
- [tex]\( P(\text{Stopped with seat-belt}) = 0.004 \)[/tex]
- [tex]\( P(\text{Stopped without seat-belt}) = 0.003 \)[/tex]
2. Calculate the total probability of being stopped:
[tex]\[ P(\text{Stopped}) = P(\text{Seat-belt on}) \times P(\text{Stopped with seat-belt}) + P(\text{No seat-belt}) \times P(\text{Stopped without seat-belt}) \][/tex]
[tex]\[ P(\text{Stopped}) = 0.61 \times 0.004 + 0.39 \times 0.003 = 0.00244 + 0.00117 = 0.00361 \][/tex]
3. Calculate the probability of having a seat-belt on when stopped:
[tex]\[ P(\text{Seat-belt on} \mid \text{Stopped}) = \frac{P(\text{Seat-belt on}) \times P(\text{Stopped with seat-belt})}{P(\text{Stopped})} \][/tex]
[tex]\[ P(\text{Seat-belt on} \mid \text{Stopped}) = \frac{0.61 \times 0.004}{0.00361} = \frac{0.00244}{0.00361} \approx 0.676 \][/tex]
Therefore, the probability that a stopped driver will have a seat-belt on is approximately [tex]\( 0.676 \)[/tex].
#### Part (b): Probability that a stopped driver will not have a seat-belt on.
1. Calculate the probability of not having a seat-belt on when stopped:
[tex]\[ P(\text{No seat-belt} \mid \text{Stopped}) = \frac{P(\text{No seat-belt}) \times P(\text{Stopped without seat-belt})}{P(\text{Stopped})} \][/tex]
[tex]\[ P(\text{No seat-belt} \mid \text{Stopped}) = \frac{0.39 \times 0.003}{0.00361} = \frac{0.00117}{0.00361} \approx 0.324 \][/tex]
Therefore, the probability that a stopped driver will not have a seat-belt on is approximately [tex]\( 0.324 \)[/tex].
### Summary:
- (a) The probability that a stopped driver will have a seat-belt on is approximately [tex]\( 0.676 \)[/tex].
- (b) The probability that a stopped driver will not have a seat-belt on is approximately [tex]\( 0.324 \)[/tex].
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