Miscellaneous · Q1: $P(B\mid A) = 1$ if $A \subset B$⁠, and $0$ if $A \cap B = \phi$

Basics you need

Part (i): A is a subset of B

1. What $A \subset B$ tells us

   Every outcome of $A$ is also an outcome of $B$⁠. So the outcomes common to $A$ and $B$ are just the outcomes of $A$⁠:

   $$A \cap B = A, \quad\text{so}\quad P(A \cap B) = P(A).$$
   The definition of $P(B|A)$ needs $P(A \cap B)$⁠. The set relation in the question tells us what $A \cap B$ is, without any numbers.
   Events are sets of outcomes. Basics: events as sets

2. Apply the definition

   Since $P(A) \neq 0$⁠, we may divide by it:

   $$\begin{aligned} P(B|A) &= \frac{P(A \cap B)}{P(A)} \\ &= \frac{P(A)}{P(A)} = 1. \end{aligned}$$
   Once we know $A$ has happened, the outcome is one of the outcomes of $A$⁠, and each of these is in $B$⁠. So $B$ is certain, and a certain event has probability 1. This is the same idea as Property 1 of Section 13.2.1, $P(F|F) = 1$⁠.
   $P(E|F) = \frac{P(E \cap F)}{P(F)}$⁠, $P(F) \neq 0$⁠. Lesson: conditional probability

Part (ii): $A \cap B = \phi$

1. What $A \cap B = \phi$ tells us

   $A$ and $B$ have no outcome in common, so their intersection is the impossible event:

   $$P(A \cap B) = P(\phi) = 0.$$
   Events with $A \cap B = \phi$ are called mutually exclusive: they cannot both happen. The empty event contains no outcomes, so its probability is 0.
   $P(\phi) = 0$⁠. Basics: rules of probability

2. Apply the definition

   $$\begin{aligned} P(B|A) &= \frac{P(A \cap B)}{P(A)} \\ &= \frac{0}{P(A)} = 0. \end{aligned}$$
   0 divided by a nonzero number is 0. In words: once we know $A$ has happened, $B$ cannot have happened as well, so its conditional probability is 0, however likely $B$ was to begin with.
   $P(E|F) = \frac{P(E \cap F)}{P(F)}$⁠, $P(F) \neq 0$⁠. Lesson: conditional probability

3. Test both answers on a die

   Throw a fair die once, $S = \{1, 2, 3, 4, 5, 6\}$⁠.

   • Part (i): $A = \{2\}$ and $B = \{2, 4, 6\}$⁠, so $A \subset B$⁠. Then $A \cap B = \{2\}$ and $P(B|A) = \frac16 \div \frac16 = 1$⁠.
   • Part (ii): $A = \{1, 3\}$ and $B = \{2, 4\}$⁠, so $A \cap B = \phi$⁠. Then $P(B|A) = 0 \div \frac26 = 0$⁠.
   An example cannot prove a general result, but it shows what the two answers mean: knowing the die shows 2 makes "even" certain, and knowing it shows 1 or 3 rules out 2 or 4.
   $P(E) = \frac{n(E)}{n(S)}$ for equally likely outcomes. Basics: classical probability

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