Question

score on last try: 0 of 1 pts. see details for more. > next question get a similar question you can retry this question below the table summarizes results from 984 pedestrian deaths that were caused by automobile accidents. pedestrian deaths driver intoxicated? pedestrian intoxicated? yes no yes 60 69 no 252 603 if one of the pedestrian deaths is randomly selected, find the probability that the pedestrian was intoxicated or the driver was not intoxicated. please enter a decimal to 4 decimal places. probability = 0.9391 ×

Explanation:

Step1: Identify relevant values

Let \(A\) be the event that the pedestrian was intoxicated and \(B\) be the event that the driver was not intoxicated.
The total number of pedestrian - deaths \(n = 984\).
The number of pedestrians intoxicated \(n(A)=60 + 252=312\).
The number of drivers not intoxicated \(n(B)=252+603 = 855\).
The number of pedestrians intoxicated and drivers not intoxicated \(n(A\cap B)=252\).

Step2: Use the addition - rule for probability

The addition - rule for probability is \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\).
We know that \(P(A)=\frac{n(A)}{n}=\frac{312}{984}\), \(P(B)=\frac{n(B)}{n}=\frac{855}{984}\), and \(P(A\cap B)=\frac{n(A\cap B)}{n}=\frac{252}{984}\).
\[

$$\begin{align*} P(A\cup B)&=\frac{312 + 855- 252}{984}\\ &=\frac{915}{984}\\ &\approx0.9390 \end{align*}$$

\]

Answer:

\(0.9390\)