Question
Asked Dec 3, 2019
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In a certain year, 27.6% of all light vehicles were light trucks, 27.6% were SUVs, and 44.8% were cars. The probability that a severe side-impact crash would prove deadly to a driver depended on the type of vehicle he or she was driving at the time, as shown in the table. What is the probability that the victim of a deadly side-impact accident was driving an SUV? (Round your answer to four decimal places.)
 

Light Truck .210
SUV .371
Car 1.000
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Expert Answer

Step 1

The probability that the victim of a deadly side-impact accident was driving an SUV is obtained below:

Let A be the event of driving Light truck, B be the event of selecting SUV, C be the event of selecting Car.

Hence the probability of selecting any group is equally likely and mutually exclusive.

That is, P (A) = 0.276, P(B) = 0.276  and P(C) = 0.448.

Let S be the event that the selected deadly proved accident occurred to the driver, then the priory probability are,

P(S|A)= 0.210, P(S|B) = 0.371, and P(S|C) = 1.000

The deadly proved accident occurred to driver is

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P(S) P(A) P(SIA)+P(B) P(S / B) + P (C) P(SIC) 0.276 (0.210)+0.276 (0.371) + (0.448) (1.000) 0.05796 0.1024+0.448 0.6084

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Step 2

The probability that the victim of a deadly sid...

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P(B) P(S B) P(A) P(SIA)+P(B) P(S B)+P(C) P(S/C) P(B|S) 0.1024 0.276(0.210)+0.276(0.371) +(0.448) (1.000) 0.1024 0.6084 =0.1683

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