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 xx is a total number of accidents. 0.2x0.2x is a number of accidents involving bodily injury. 0.2x0.4=0.08x0.2x\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(AB)=P(A)P(B)P(A\cap B)=P(A)P(B) is equivalent to independency of events AA and BB (see [1, page 125]). Event AA : The accident was partly caused by weather conditions; Event BB: The accident involved bodily injury. P(AB)=0.20.4=0.08,P(A\cap B)=0.2\cdot0.4=0.08, whereas, P(A)P(B)=0.20.3=0.06P(A)P(B)=0.2\cdot0.3=0.06. Thus, the events are not independent.

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

P(BA)=P(AB)P(A)=0.20.40.3=0.2667P(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% (0.20.8100%)(0.2\cdot0.8\cdot100\%) of accidents were partly caused by weather conditions and involved bodily injuries. Thus, 100%30%20%+8%=58%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.


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