9/17/2014 Using observations made of drivers arriving at a certain high school, a study reported that 69% of adults wear seat-belts while driving. A high school student also in the car wears a seat-belt 66% of the time when the adult wears a seat-belt, and 26% of the time when the adult does not wear a seat-belt.
a) Write the given probabilities using Notation. Make a table to depict the problem.
b) What is the probability that a high school student in the study wears a seat-belt?
c) A car is randomly selected and found to contain a student wearing his seat-belt. What is the probability that the adult driver was wearing a seat-belt?
d) What is the probability that at least 1 of them wears a seat-belt?
e) What is the probability that only 1 of them wears a seat-belt?

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9/17/2014 Solution.
a) Given: P(ASB) = 0.69P(ASB') = 0.31;
P(SSB | ASB) = 0.66; P(SSB | ASB') = 0.26

 ASB ASB' SSB 0.4554 0.0806 0.536 SSB' 0.2346 0.2294 0.464 0.69 0.31 1

So, P(ASB and SSB) = P(SSB | ASB)*P(ASB) = 0.69*0.66 = 0.4554
P(ASB' and SSB) = P(SSB | ASB')*P(ASB') = 0.31*0.26 = 0.0806

b) P(SSB) = P(ASB and SSB) + P(ASB' and SSB) = 0.4554 + 0.0806 = 0.536

c) P(ASB | SSB) = P(ASB and SSB) / P(SSB) = 0.4554/0.536 = 0.8496
d) P(at least 1 wears a seatbelt) = 1 – P(neither does) = 1 – P(ASB' and SSB') = 1 – 0.2294 = 0.7706
e) P(only 1 wears a seat-belt) = P(ASB' and SSB) + P(ASB and SSB') =
0.0806 + 0.2346 = 0.3152