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
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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