Math Home

Probability

Claim: Let \(\Omega\) be a sample space with a probability function \(P\) on an event space \(\mathcal{E}.\) Let \(F\) be an event such that \(P(F) > 0.\) Then the function \[P(\cdot | F): \mathcal{E} \rightarrow [0, 1]\] is a probability.

▼ Proof:

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}

Example: You are given the following table of information about a survey given to test ice cream flavors:

Likes Lemon Only

Likes Berry Only

Likes Both

Adult

\(5\%\)

\(34\%\)

\(32\%\)

\(5\%\)

\(34\%\)

\(32\%\)

Child

\(3\%\)

\(77\%\)

\(16\%\)

\(3\%\)

\(77\%\)

\(16\%\)

What is the probability that an adult likes at least one of the flavors?

▼ Solution:

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.

Claim: The function \(P(F | \cdot)\) is not a probability.

▼ Counterexample:

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, \(P(F|A) = P(F|B) = 2 \neq P(F|A \cup B).\)

1. The events \(A\) and \(B\) satisfy the following:
\[P(A) = 0.5, P(B) = 0.4, P(A|B) = 0.8.\]
Find \(P(A^C|B).\)

Unanswered

2. If \(P(A|C) = 0.4,\) \(P(B|C) = 0.6,\) and \(P(A \cap B|C) = 0.1,\) what is \(P(A \cup B|C)?\)

Unanswered