ABy Admin
Apr 29'23
Exercise
A study of automobile accidents produced the following data:
Model year | Proportion of all vehicles | Probability of involvement in an accident |
---|---|---|
2014 | 0.16 | 0.05 |
2013 | 0.18 | 0.02 |
2012 | 0.2 | 0.03 |
Other | 0.46 | 0.04 |
An automobile from one of the model years 2014, 2013, and 2012 was involved in an accident. Calculate the probability that the model year of this automobile is 2014.
- 0.22
- 0.30
- 0.33
- 0.45
- 0.50
ABy Admin
Apr 29'23
Solution: D
Let B, C, and D be the events of an accident occurring in 2014, 2013, and 2012, respectively. Let A = B ∪ C ∪ D .
[[math]]
\operatorname{P}[B | A] = \frac{P[ A | B]P[ B]}{P[ A | B][ P[ B] + P[ A | C ]P[C ] + P[ A | D]P[ D]}
[[/math]]
Use Bayes’ Theorem
[[math]]
\frac{(0.05)(0.16)}{(0.05)(0.16) + (0.02)(0.18) + (0.03)(0.20)} = 0.45.
[[/math]]