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?
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
seat-belt) = P(ASB' and SSB) + P(ASB and SSB') =
0.0806 + 0.2346 = 0.3152