Two pumps, A and B, are responsible for supplying water to a town. The probability that pump A fails is 0.1, and the probability that pump B fails is 0.2. The pumps operate independently. Calculate: a

The probability that both pumps are working is $\boxed{0.72}$.

The probability that there is no water in the town is $\boxed{0.02}$.

• Pump A is working AND pump B is failing.
• Pump A is failing AND pump B is working.

• $P(A_w \text{ and } B_f) = P(A_w) \times P(B_f) = 0.9 \times 0.2 = 0.18$
• $P(A_f \text{ and } B_w) = P(A_f) \times P(B_w) = 0.1 \times 0.8 = 0.08$

The probability that only one pump is working is $\boxed{0.26}$.

This is the complement of "no water in the town".

$P(\text{some water}) = 1 - P(\text{no water})$.

Alternatively, we can sum the probabilities of the cases where at least one pump is working:
$P(\text{some water}) = P(\text{both working}) + P(\text{only A working}) + P(\text{only B working})$
$P(\text{some water}) = P(\text{both working}) + P(A_w \text{ and } B_f) + P(A_f \text{ and } B_w)$
$P(\text{some water}) = 0.72 + 0.18 + 0.08$
$P(\text{some water}) = 0.98$
The probability that there is some water in the town is $\boxed{0.98}$.