Probability; simple homework. Cars with or without functioning lights or brakes The Task: During a technical control lights and brakes was controlled on randomly selected cars. 18% had errors with the lights and 12% with the brakes. 74% had both lights and brakes in order. What is the probability that:
a) The lights were in order?
b) The brakes were in order?
c) No brakes and no lights?
d) Lights, but no brakes.  
My answers(d is the problem):
a) 82%
b) 88%
c) 18+12-26=4%
d) What I have scribbled so far:
p(A) = Lights = 0.88, p(B) = brakes = 0.88 $P(A \cap B) = 0.74$
$P(A \cup B) = 0.88+0.82 - 0.74 = 0.96$ At least one of the two is functioning.
The complement of that last would be both is not working; 4%.
26% that both or at least one is not working.  
From there I don't see what I'm suppose to do. The book says the answer is 8% on d). I'll check back here in the morning. Good night. 
 A: Please draw a Venn Diagram. Then everything will become clear.
I don't know how to draw a picture on this site, so will use $1000$ words instead.
Draw a rectangle.  The rectangle (including the points inside) symbolically represents all cars, meaning $100\%$.
Draw two intersecting ovals inside the rectangle.  The left oval is all no light cars, the right oval is all no brake cars.  The region of intersection of the two ovals is then the no light no brake cars.  Maybe label the left oval NL, the right oval NB. 
[Or else the two ovals could represent good light, good brake. Doesn't really matter too much. But I will stick to my first choice, so forget this paragraph.]
Region NL has (symbolically) size $18$ (will forget about $\%$ stuff temporarily). Region NB has size $12$.  But you saw that altogether they cover a region of size  $26$ only, $4$ down from the sum $18+12$.  So the intersection of the two ovals must represent size $4$. Region NL and NB has therefore size $4$, our probability is $4\%$.
Finally, we look at part (d), light and no brake.  That's the region in NB but not in (NL and NB).  Make sure you are clear on what region this is.
Its size is $12-4$. This is because NB has size $12$, but the common region has size $4$.  Thus the correct answer is $8\%$.  
All these calculations can be expressed using more formal language that involves $\cup$, $\cap$, and complement.  But in my opinion the picture comes first, to supply the intuition.  Once you know what's really happening, expressing this knowledge in set-theoretic language is fairly mechanical.
A: If you know that 74% have both lights and brakes working and 82% have lights working, then 82%-74%=8% have lights but no brakes.  Similarly 88%-74%=14% have brakes but no lights.  That means 100%-74%-8%-14%=4% have no lights and no brakes.
A: Hint: 82%-74%=8%
good luck,you just try with (no error with the light- 74%)!
