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Sagot :
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### Given Data:
- Red Car with Speeding Ticket: 15
- Red Car without Speeding Ticket: 135
- Not Red Car with Speeding Ticket: 45
- Not Red Car without Speeding Ticket: 470
### Total number of people:
To find the total number of people surveyed, sum up all the values in the table:
[tex]\[ \text{Total People} = 15 + 135 + 45 + 470 = 665 \][/tex]
### Step-by-Step Calculation:
#### Part (a): Probability of a Speeding Ticket or Having a Red Car
We need to find the probability of a person either having a speeding ticket or having a red car.
1. Probability of having a Speeding Ticket:
[tex]\[ P(\text{Speeding Ticket}) = \frac{\text{Number with Speeding Ticket}}{\text{Total People}} = \frac{15 + 45}{665} = \frac{60}{665} \][/tex]
2. Probability of having a Red Car:
[tex]\[ P(\text{Red Car}) = \frac{\text{Number of Red Cars}}{\text{Total People}} = \frac{15 + 135}{665} = \frac{150}{665} \][/tex]
3. Probability of having both a Speeding Ticket and a Red Car (since we counted these people twice above):
[tex]\[ P(\text{Speeding Ticket and Red Car}) = \frac{15}{665} \][/tex]
4. Combined Probability (using the formula for the union of two events):
[tex]\[ P(\text{Speeding Ticket or Red Car}) = P(\text{Speeding Ticket}) + P(\text{Red Car}) - P(\text{Speeding Ticket and Red Car}) \][/tex]
Plugging in the numbers:
[tex]\[ P(\text{Speeding Ticket or Red Car}) = \frac{60}{665} + \frac{150}{665} - \frac{15}{665} = \frac{195}{665} - \frac{15}{665} = \frac{180}{665} \][/tex]
[tex]\[ P(\text{Speeding Ticket or Red Car}) \approx 0.2932 \][/tex]
Thus, the probability that a randomly chosen person has either a speeding ticket or a red car is approximately 0.2932.
#### Part (b): Probability that Two Randomly Chosen People Both Have a Red Car
We need to find the probability that two randomly chosen people both have a red car.
1. Probability that the first person has a Red Car:
[tex]\[ P(\text{First person has a Red Car}) = \frac{150}{665} \][/tex]
2. Probability that the second person, given the first is already chosen, also has a Red Car:
If the first person chosen has a red car, then there are [tex]\(150 - 1 = 149\)[/tex] red cars left out of [tex]\(665 - 1 = 664\)[/tex] people left.
[tex]\[ P(\text{Second person has a Red Car | first person has a Red Car}) = \frac{149}{664} \][/tex]
3. Combined Probability:
[tex]\[ P(\text{Both people have a Red Car}) = \frac{150}{665} \times \frac{149}{664} \][/tex]
[tex]\[ P(\text{Both people have a Red Car}) \approx 0.0505 \][/tex]
Thus, the probability that two randomly chosen people both have a red car is approximately 0.0505.
### Given Data:
- Red Car with Speeding Ticket: 15
- Red Car without Speeding Ticket: 135
- Not Red Car with Speeding Ticket: 45
- Not Red Car without Speeding Ticket: 470
### Total number of people:
To find the total number of people surveyed, sum up all the values in the table:
[tex]\[ \text{Total People} = 15 + 135 + 45 + 470 = 665 \][/tex]
### Step-by-Step Calculation:
#### Part (a): Probability of a Speeding Ticket or Having a Red Car
We need to find the probability of a person either having a speeding ticket or having a red car.
1. Probability of having a Speeding Ticket:
[tex]\[ P(\text{Speeding Ticket}) = \frac{\text{Number with Speeding Ticket}}{\text{Total People}} = \frac{15 + 45}{665} = \frac{60}{665} \][/tex]
2. Probability of having a Red Car:
[tex]\[ P(\text{Red Car}) = \frac{\text{Number of Red Cars}}{\text{Total People}} = \frac{15 + 135}{665} = \frac{150}{665} \][/tex]
3. Probability of having both a Speeding Ticket and a Red Car (since we counted these people twice above):
[tex]\[ P(\text{Speeding Ticket and Red Car}) = \frac{15}{665} \][/tex]
4. Combined Probability (using the formula for the union of two events):
[tex]\[ P(\text{Speeding Ticket or Red Car}) = P(\text{Speeding Ticket}) + P(\text{Red Car}) - P(\text{Speeding Ticket and Red Car}) \][/tex]
Plugging in the numbers:
[tex]\[ P(\text{Speeding Ticket or Red Car}) = \frac{60}{665} + \frac{150}{665} - \frac{15}{665} = \frac{195}{665} - \frac{15}{665} = \frac{180}{665} \][/tex]
[tex]\[ P(\text{Speeding Ticket or Red Car}) \approx 0.2932 \][/tex]
Thus, the probability that a randomly chosen person has either a speeding ticket or a red car is approximately 0.2932.
#### Part (b): Probability that Two Randomly Chosen People Both Have a Red Car
We need to find the probability that two randomly chosen people both have a red car.
1. Probability that the first person has a Red Car:
[tex]\[ P(\text{First person has a Red Car}) = \frac{150}{665} \][/tex]
2. Probability that the second person, given the first is already chosen, also has a Red Car:
If the first person chosen has a red car, then there are [tex]\(150 - 1 = 149\)[/tex] red cars left out of [tex]\(665 - 1 = 664\)[/tex] people left.
[tex]\[ P(\text{Second person has a Red Car | first person has a Red Car}) = \frac{149}{664} \][/tex]
3. Combined Probability:
[tex]\[ P(\text{Both people have a Red Car}) = \frac{150}{665} \times \frac{149}{664} \][/tex]
[tex]\[ P(\text{Both people have a Red Car}) \approx 0.0505 \][/tex]
Thus, the probability that two randomly chosen people both have a red car is approximately 0.0505.
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