To show that this claim is true, we need to show that \(P(\cdot|F)\) satisfies the axioms of a probability.

First check that \(P(\Omega|F) = 1.\) \begin{align} P(\Omega | F) & = \frac{P(\Omega \cap F)}{P(F)} \\ & = \frac{P(F)}{P(F)} \\ & = 1 \end{align}

Next, check that if \(A_1, A_2, \dots\) are disjoint events in \(\mathcal{E}\) then \[P\left(\left.\bigcup_{i=1}^\infty A_i \right| F\right) = \sum_{i=1}^\infty P(A_i | F)\] Start with the left hand side: \[P\left(\left.\bigcup_{i=1}^\infty A_i \right| F\right) = \frac{P\left(\left(\bigcup_{i=1}^\infty A_i \right) \cap F\right)}{P(F)}\] Since \(\left(\bigcup_{i=1}^\infty A_i \right) \cap F = \bigcup_{i=1}^\infty (A_i \cap F),\) we have \[\frac{P\left(\left(\bigcup_{i=1}^\infty A_i \right) \cap F\right)}{P(F)} = \frac{P\left(\bigcup_{i=1}^\infty (A_i \cap F)\right)}{P(F)}\] Also, since \(A_1, A_2, \dots\) are disjoint, so are \(A_1 \cap F, A_2 \cap F, \dots,\) so \begin{align} \frac{P\left(\bigcup_{i=1}^\infty (A_i \cap F)\right)}{P(F)} & = \frac{\sum_{i=1}^\infty P(A_i \cap F)}{P(F)} \\ & = \sum_{i=1}^\infty \frac{P(A_i \cap F)}{P(F)} \\ & = \sum_{i=1}^\infty P(A_i|F) \end{align}

Let \(L\) be the event that a person likes lemon only, \(B\) be the event that a person likes berry only, \(E\) be the event that a person likes both flavors, and \(A\) be the event that a person is an adult.

The event that an adult likes at least one of the flavors is \(P(L \cup B \cup E|A).\) Since \(L,\) \(B\) and \(E\) are disjoint events and \(P(\cdot|A)\) is a probability, \[P(L \cup B \cup E|A) = P(L|A) + P(B|A) + P(E|A)\] From the table we see \(P(L|A) = 0.05,\) \(P(B|A) = 0.34\) and \(P(E|A) = 0.32.\) So, \[P(L|A) + P(B|A) + P(E|A) = 0.05 + 0.34 + 0.32 = 0.71\] There is a \(71\%\) chance that an adult will like at least one of the ice cream flavors.

For example, consider the outcome of two coin flips. Let \(F\) be the event that the first flip is a heads. So, \(F = \{HH, HT\}.\) Let \(A\) be the event that both flips are heads, \(A = \{HH\}.\) Let \(B\) be the event that the first flip is heads and the second flip is tails, \(B = \{HT\}.\)

The sets \(A\) and \(B\) are disjoint, so if \(P(F | \cdot)\) were a probability then \(P(F | A \cup B) = P(F|A) + P(F|B).\) However, \(P(F|A \cup B) = P(F|A) = P(F|B) = 1.\) So, \(P(F|A) = P(F|B) = 2 \neq P(F|A \cup B).\)