Let event
A=pump A is working
B=pump B is working
Assume pumps A and B are independent.
then
A'=pump A is not working
B'=pump B is not working,
hence
P(A')=0.1 (probability that pump A fails)
P(A)=1-0.1=0.9 (probaility that pump A works)
P(B')=0.2
P(B)=1-0.2=0.8
(a) Both pumps are working
use the multiplication rule for both steps to succeed.
P(A∩B)=P(A)*P(B)
(b) Both pumps are not working
similar to (a), but both steps will fail.
P(A'∩B')
(c) Only one pump is working
P(A∩B')+P(A'∩B)
If A works, then B doesn't and vice versa.
(d) At least one pump is working:
subtract result (b) from 1, i.e. any case but two pumps failing at the same time.
The water supply in a town depends entirely on two pumps. A and B. The probability of pump A filling is 0.1 and the probability of pump B failing is 0.2. Calculate the probability that
(a) Both pumps are working
(b) There is no water in the town
(c) Only one pump is working
(d) There is some water in the town
1 answer