Answer to Question #126783 in Statistics and Probability for jaya

Question #126783
c) An insurance company estimated that 30% of all automobile accidents were partly caused by
weather conditions and that 20% of all automobile accidents involved bodily injury. Further, of
those accidents that involved bodily injury, 40% were partly caused by weather conditions.
I. What is the probability that a randomly chosen accident both was partly caused by
weather conditions and involved bodily injury?
II. Are the events “partly caused by weather conditions” and “involved bodily injury”
independent?
III. If a randomly chosen accident was partly caused by weather conditions, what is the
probability that it involved bodily injury?
IV. What is the probability that a randomly chosen accident both was not partly caused by
weather conditions and did not involve bodily injury?
1
Expert's answer
2020-07-23T18:30:41-0400

i. Suppose that "x" is a total number of accidents. "0.2x" is a number of accidents involving bodily injury. "0.2x\\cdot0.4=0.08x" is a number of accidents involving bodily injury that were caused by weather conditions. Thus, the probability is 0.08.

ii. The following condition: "P(A\\cap B)=P(A)P(B)" is equivalent to independency of events "A" and "B" (see [1, page 125]). Event "A" : The accident was partly caused by weather conditions; Event "B": The accident involved bodily injury. "P(A\\cap B)=0.2\\cdot0.4=0.08," whereas, "P(A)P(B)=0.2\\cdot0.3=0.06". Thus, the events are not independent.

iii. Our aim is to compute "P(B|A)" . We remind that (see [1, page 115])

"P(B|A)=\\frac{P(A\\cap B)}{P(A)}=\\frac{0.2\\cdot0.4}{0.3}=0.2667".

iv. We know that 30% of accidents were partly caused by weather conditions and 20% of accidents involved bodily injuries. 8% "(0.2\\cdot0.8\\cdot100\\%)" of accidents were partly caused by weather conditions and involved bodily injuries. Thus, "100\\%-30\\%-20\\%+8\\%=58\\%" of accidents were not partly caused by weather conditions and did not involve bodily injuries. Thus, the respective probability is 0.58

References:

[1] Feller, W. (1967). An Introduction to Probability Theory and Its Applications (Third ed.). New York: Wiley.

Answers: i. 0.08; ii. The events are not independent; iii. 0.2667; iv. 0.58.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS