Math  /  Data & Statistics

QuestionBy examining the past driving records of drivers in a certain city, an insurance company has determined the (empirical) probabilities in the table to the right. Use the (empirical) probabilities to complete parts (A)(A) and (B)(B) below. \begin{tabular}{|c|c|c|c|} \hline \multirow[t]{2}{*}{} & \multicolumn{3}{|l|}{Miles Driven per Year} \\ \hline &  Less  Than 10,009M1\begin{array}{c} \hline \text { Less } \\ \text { Than } \\ 10,009 \\ \mathrm{M}_{1} \end{array} & \begin{tabular}{l} 10,00015,000\begin{array}{c} 10,000- \\ 15,000 \end{array} \\ Inclusive, M2M_{2} \end{tabular} &  More  Than 15,000M3\begin{array}{c} \text { More } \\ \text { Than } \\ 15,000 \text {, } \\ \mathrm{M}_{3} \end{array} \\ \hline Accident A & 0.10 & 0.20 & 0.20 \\ \hline No Accident A\mathrm{A}^{\prime} & 0.05 & 0.20 & 0.25 \\ \hline Totals & 0.15 & 0.40 & 0.45 \\ \hline \end{tabular} (A) Find the probability that a city driver selected at random drives more than 15,000 miles per year or has an accident.
The probability is 0.75 . (Type an integer or a decimal.) (B) Find the probability that a city driver selected at random drives 15,000 or fewer miles per year and has an accident.
The probability is \square (Type an integer or a decimal.)

Studdy Solution

STEP 1

1. The table provides empirical probabilities for different categories of miles driven and accident occurrences.
2. We assume the probabilities in each category are mutually exclusive and collectively exhaustive.

STEP 2

1. Solve part (A): Calculate the probability that a driver drives more than 15,000 miles per year or has an accident.
2. Solve part (B): Calculate the probability that a driver drives 15,000 or fewer miles per year and has an accident.

STEP 3

To find the probability that a driver drives more than 15,000 miles per year or has an accident, use the formula for the probability of the union of two events:
P(M3A)=P(M3)+P(A)P(M3A) P(M_3 \cup A) = P(M_3) + P(A) - P(M_3 \cap A)
Where: - P(M3) P(M_3) is the probability of driving more than 15,000 miles. - P(A) P(A) is the probability of having an accident. - P(M3A) P(M_3 \cap A) is the probability of both driving more than 15,000 miles and having an accident.

STEP 4

Calculate each component:
- P(M3)=0.45 P(M_3) = 0.45 - P(A)=0.10+0.20+0.20=0.50 P(A) = 0.10 + 0.20 + 0.20 = 0.50 - P(M3A)=0.20 P(M_3 \cap A) = 0.20

STEP 5

Substitute the values into the formula:
P(M3A)=0.45+0.500.20=0.75 P(M_3 \cup A) = 0.45 + 0.50 - 0.20 = 0.75

STEP 6

To find the probability that a driver drives 15,000 or fewer miles per year and has an accident, consider the events M1 M_1 and M2 M_2 for miles driven:
P((M1M2)A)=P(M1A)+P(M2A) P((M_1 \cup M_2) \cap A) = P(M_1 \cap A) + P(M_2 \cap A)
Where: - P(M1A)=0.10 P(M_1 \cap A) = 0.10 - P(M2A)=0.20 P(M_2 \cap A) = 0.20

STEP 7

Calculate the probability:
P((M1M2)A)=0.10+0.20=0.30 P((M_1 \cup M_2) \cap A) = 0.10 + 0.20 = 0.30
The probability for part (A) is:
0.75 \boxed{0.75}
The probability for part (B) is:
0.30 \boxed{0.30}

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