Question 1137226
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Define two events 
A = owning a dog
B = owning a cat


Given info:
P(A) = 0.4
P(B) = 0.3
P(A and B) = 0.15


Use the given info to compute the probability of selecting a household that has either pet
P(A or B) = P(A) + P(B) - P(A and B)
P(A or B) = 0.4 + 0.3 - 0.15
P(A or B) = 0.7 - 0.15
P(A or B) = 0.55
The probability of having either a cat, a dog, or both is 0.55


Subtract that result from 1 to find the probability of having neither pet
P(not A and not B) = 1 - P(A or B)
P(not A and not B) = 1 - 0.55
P(not A and not B) = 0.45
so 45% of the population of households has neither pet


Answer: <font color=red size=4>0.45</font>
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