Calculating combined probabiltiy

Question

Given the following:

$P(A) = 0.40$
$P(B) = 0.55$
$P(C) = 0.70$
$P(A∪B) = 0.63$
$P(A∪C) = 0.77$
$P(B∪C) = 0.80$
$P(A∪B∪C) = 0.85$

I need to find

$P(A'∩B'∩C)$

I have attempted to do this via the following math

$P(A'∩B'∩C) = P(C) - P(A∩B) - P(B∩C) + P(A∩B∩C)$

$P(A∩B) = P(A) + P(B) - P(A∪B) = 0.4+0.55-0.63 = 0.32$

$P(B∩C) = P(B) + P(C) - P(B∪C) = 0.55+0.7-0.8 = 0.45$

$P(A∩B∩C) = P(A)+P(B)+P(C) - P(A∪B∪C) = 0.4+0.55+0.7-0.85 = 0.8$

So that leaves me with

$0.7 - 0.32 - 0.45 + 0.8 = 0.73$, but the answer should be $0.22$

I suspect my error lies within the $P(A∩B∩C)$ solution. What am I doing wrong?

Answer 1

$$P(A' \cap B' \cap C)+P(A \cap B \cap C) + P(A \cap B' \cap C) + P(A' \cap B \cap C) = P(C)$$

$$P(A' \cap B' \cap C) + P(A \cap B' \cap C) + P( B \cap C) = P(C)$$

$$P(A' \cap B' \cap C) = P(C)-P(B \cap C)- P(A \cap B' \cap C) $$

$$P(A' \cap B' \cap C) = P(C)-P(B \cap C)- P(A \cap C) +P(A \cap B \cap C)$$

Your formula is

$$P(A'\cap B' \cap C)=P(C)-P(B \cap C)+P(A\cap B\cap C)-P(A \cap \color{red}B)$$

and as pointed out in the comment, the formula for $P(A \cap B \cap C)$ is wrong as well.

Answer 2

None of the given probabilities are for terms containing complements, so I suggest beginning with deMorgan's Law for Sets, and use probability for the relative complement, thus eliminating the complementations as early as possible.   Otherwise you just confuse yourself with the jumble.$$\mathsf P(A'\cap B'\cap C) {=\mathsf P((A\cup B)'\cap C)\\ = \mathsf P(C)-\mathsf P((A\cup B)\cap C)\\ = \mathsf P(C)-\mathsf P(A\cup B)-\mathsf P(C)+\mathsf P(A\cup B\cup C)\\\vdots\\=0.22}$$