9/17/2014
Using
observations made of drivers arriving at a certain high school, a
study reported that 69% of adults wear seatbelts while driving. A
high school student also in the car wears a seatbelt 66% of the time
when the adult wears a seatbelt, and 26% of the time when the adult
does not wear a seatbelt.
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
seatbelt?
c) A car is randomly selected and found to contain a
student wearing his seatbelt. What is the probability that the adult
driver was wearing a seatbelt?
d) What is the probability that at
least 1 of them wears a seatbelt?
e) What is the probability that
only 1 of them wears a seatbelt?
Scroll down for the solution.
9/17/2014
Solution.
a)
Given: P(ASB)
= 0.69
→ P(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
seatbelt) = P(ASB' and SSB) + P(ASB and SSB') =
0.0806 + 0.2346 = 0.3152