In arbitrary town 50% families have a dog (A), 30% cat (B), 10% fish (C), 20% dog and cat, 8% dog and fish, 5% cat and fish, 3% all of them.
What is the probability:
- that randomly chosen family has none of the animals
- that randomly chosen family has exactly one of the animals
First is easy: $$ P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(A \cap C) + P(A \cap B \cap C) = \dots = \frac{3}{5} $$ so $P(A^C \cap B^C \cap C^C) = 1 - P(A \cup B \cup C) = 1 - \frac{3}{5} = \frac{2}{5}$.
I am however, stuck on the second question. How can I solve this?
A - Dog, B - Cat and C-Fish
$P$(A only) = $P(A)-P(A\cap B) - P(A\cap C) +P(A\cap B\cap C) =0.5-0.2-.08+0.03 = 0.25$
$P$(B only) = $P(B)-P(A\cap B) - P(B\cap C) +P(A\cap B\cap C) = 0.3-0.2-0.05+.03 = 0.08$
$P$(C only) = $P(C)-P(C\cap B) - P(A\cap C) +P(A\cap B\cap C)= 0.1-0.08-0.05+0.03 = 0$
$P$( exactly one animal) = $ 0.25+0.08+0.0 = 0.33$