What is the probability that one of your customers does not sign up either way?

Question

Is my math correct for this question?

Suppose that $25\%$ of your customers sign up for special offers via email, and $20\%$ of your customers sign up for special offers via regular mail. Moreover, $5\%$ of all your customers sign up for both. What is the probability that one of your customers does not sign up either way?

$1-P(A)+P(B)-P(A\cap B)=P(\text{not }A \cup \text{not }B)$
$1-.25+.2-.05 = .6 = 60\%$

I'm pretty sure it's correct, but the online hw is marking it wrong, could use suggestions if they're wrong.

Answer 1

There is miscalculation $=1-0.25+0.2-0.05=0.9\neq0.6$

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Probability of neither $A$ nor $B$ :

$P(\overline{A}\cap\overline{B})=1-P\overline{(A\cup B)}$

$=1-(P(A)+P(B)-P(A \cap B))$

$=1-P(A)-P(B)+P(A\cap B)$

$=1-0.25-0.20+0.05$

$=0.60$

$60\%$ of customers does not sign up either way.

Answer 2

This is wrong:

$1-P(A)+P(B)-P({A}\cap{B})=P(\neg{A}\cup\neg{B})$

This is right:

$1-\color\red(P(A)+P(B)-P({A}\cap{B})\color\red)=P(\neg{A}\cup\neg{B})$

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Using the right formula would give you $1-(0.25+0.20-0.05)=0.6$.