# 2.14 - (c Let F denote “fatal accident” We have 50...

This preview shows page 1. Sign up to view the full content.

2.14 (a) Note that the question assumes an accident either occurs or does not occur at a given crossing each year, i.e. no more than one accident per year. There were a total of 30 + 20 + 60 + 20 = 130 accidents in 10 years, hence the yearly average is 130 / 10 = 13 accidents among 1000 crossings, hence the yearly probability of accident occurring at a given crossing is 1000 13 = 0.013 (probability per year) (b) Examining the data across the “Day” row, we see that the relative likelihood of R compared to S is 30:60, hence P(S | D) = 60/90 = 2/3
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: (c) Let F denote “fatal accident”. We have 50% of (30 + 20) = 0.5 × 50 = 25 fatal “run into train” accidents, and 80% of (60 + 20) = 0.8 × 80 = 64 fatal “struck by train” accidents, hence the total is P(F) = (25 + 64) / 130 ≅ 0.685 (d) (i) D and R are not mutually exclusive; they can occur together (there were 30 run-into-train accidents happened in daytime); (ii) If D and R are, we must have P(R | D) = P(R), but here P(R | D) = 30 / 90 = 1/3, while P(R) = (30 + 20) / 130 = 5/13, so D and R are not statistically independent....
View Full Document

## This note was uploaded on 04/25/2010 for the course MAE 213 taught by Professor Seshadri,k during the Spring '08 term at UCSD.

Ask a homework question - tutors are online