An insurance company estimated that 30% of all automobile accidents were partly

Question

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. What is the probability that a randomly chosen accident both was partly caused by weather conditions and involved bodily injury? Are the events 'partly caused by weather conditions' and 'involved in bodily injury' independent? Show this mathematically. If an accident was partly caused by weather conditions, what is the probability that it involved bodily injury?

Explanation / Answer

Solution

Back-up Theory

If A and B are two events which are not independent, probability of A given that B has occurred, denoted by, P(A/B), is given by P(A/B) = P(A&B)/P(B) ……………..............………………….. (1)

From (1), P(A&B) = P(A/B) x P(B) ………………………………. ……………………(2)

(1) and (2) => P(B/A) = {P(A/B) x P(B)}/P(A) ………………………………………….(3)

If A and B are independent, P(A/B) = P(A) ……………………….……………………(4)

Now, to work out the solution,

Let event A represent ‘automobile accident is partly caused by weather conditions’ and event B represent ‘automobile accident involved bodily injury’.

We are given: P(A) = 0.3; P(B) = 0.2 and P(A/B) = 0.4

Part (a)

P(accident was partly caused by weather condition and involved bodily injury)

= P(A&B) = P(A/B) x P(B) [by (2)]

= 0.4 x 0.2 = 0.08 ANSWER

Part (b)

Here, P(A/B) is not equal to P(A) or P(B) and hence by (4),

the two events are not independent. ANSWER

Part (c)

Given an accident was caused partly by weather condition, probability it involved bodily injury

= P(B/A) = {P(A/B) x P(B)}/P(A) [by (3)]

= (0.4 x 0.2)/0.3 = 4/15 or 0.267 ANSWER

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