Outlining the Question

Let $A$ , $B$ , and $C$ be sets with dependent probabilities.

Suppose:

$\\P(A) = 0.17 \\P(B) = 0.07 \\P(C) = 0.13 \\P(A \cup B) = 0.18 \\P(A \cup C) = 0.19 \\P(B \cup C) = 0.18 \\P(A \cap B \cap C) = 0.01$

1a. Find

$P(X=0)$

Response

$P(X = 0) = P(A^c \cap B^c \cap C^c) = 1 - P(A \cup B \cup C)$

I need to calculate it

$\\P(A \cup B \cup C) = P(A) + P(B) + P(C) \\- (P(A \cap B) + P(A \cap C) + P(B \cap C)) \\+ P(A \cap B \cap C) \\= 0.17 + 0.07 + 0.13 - (P(A \cap B) + P(A \cap C) + P(B \cap C)) + 0.01$

I am still struggling to find the intersections of all of these. But I figured I could calculate the unions.

But how to isolate for any of these?

Eureka!

While tinkering I realized that the two probabilities can be extracted:

If you add up the individual probabilities and subtract the union, you end up with the intersection! Visually it looks like this:

$\\P(A \cap C^c) = P(A \cup C) - P(C) \\P(A^c \cap C) = P(A \cup C) - P(A) \\P(A \cap C) = P(A \cup C) - P(A^c \cap C) - P(A \cap C^c) \\= P(A \cup C) - P(A \cup C) + P(C) - P(A \cup C) + P(A) \\= P(A) + P(C) - P(A \cup C) \\= 0.17 + 0.13 - 0.19 = 0.11$

Updating the visual to match the simplified equation:

And now that I see it, of course it makes sense!

We can do the same for the others:

$\\P(A \cap B) = P(A) + P(B) - P(A \cup B) \\= 0.17 + 0.07 - 0.18 = 0.06$

And:

$\\P(B \cap C) = P(B) + P(C) - P(B \cup C) \\= 0.07 + 0.13 - 0.18 = 0.02$

So then:

$\\P(A \cup B \cup C) = 0.17 + 0.07 + 0.13 - 0.11 - 0.06 - 0.02 + 0.01 \\P(A \cup B \cup C) = 0.19$

And finally:

$P(X = 0) = 1 - P(A \cup B \cup C) = 1.00 - 0.19 = 0.81$

1b. Find

$P(X=1)$

Response

Now that we have isolated the intersections.

$\\P(X = 1) = P((A \cap B^c \cap C^c) \cup (A^c \cap B \cap C^c) \cup (A^c \cap B^c \cap C)) \\= P(A \cup B \cup \C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + 2 P(A \cap B \cap C) \\= 0.19 - 0.11 - 0.06 - 0.02 + 2 0.01 \\= 0.02$

1c. Find

$P(X = 2)$

Response

$\\P(X = 2) = P((A \cap B \cap C^c) \cup (A \cap B^c \cap C) \cup (A^c \cap B \cap C))\\= P(A \cap B) + P(A \cap C) + P(B \cap C) - 2 P(A \cap B \cap C) \\= 0.11 + 0.06 + 0.02 - 2 0.01 \\= 0.17$