Conditional Probability

Conditional probability measures the probability of an event when another related event is already known to have occurred. This idea helps students understand restricted sample space, the multiplication rule, independent events, and later connections to Bayes’ theorem.

If A and B are two events associated with the same sample space of a random experiment, the conditional probability of the event A given that B has occurred, i.e. P(A∣B), is given by

\[P(A | B) = \frac{P(A \cap B)}{P(B)}\], where \[P(B) \neq 0\].

\[P(B | A) = \frac{P(A \cap B)}{P(A)}\], where \[P(A) \neq 0\].

Multiplication rule: \[P(A \cap B) = P(B) \cdot P(A | B) = P(A) \cdot P(B | A)\].

Complement form: \[P(A' | B) = 1 - P(A | B)\], P(B) ≠ 0

A family has two children. What is the probability that both children are boys, given that at least one of them is a boy?

Let:

• E = both children are boys

• F = at least one child is a boy.

Step 1: Write the sample space
Using b for boy and g for girl:

These four outcomes are equally likely.

Step 2: Write the events

• E = {(b,b)}

• F = {(b,b),(g,b),(b,g)}

Step 3: Find the intersection

\[E \cap F = \{(b, b)\}\]

So,

\[P(E \cap F) = \frac{1}{4}, \quad P(F) = \frac{3}{4}\]

Step 4: Apply the formula

\[P(E | F) = \frac{P(E \cap F)}{P(F)} = \frac{1/4}{3/4} = \frac{1}{3}\]

Answer: \[\frac{1}{3}\]

• \[P(S|F) = P(F|F) = 1\]

• If A and B are any two events of a sample space S and F is an event of S such that \[P(F) \neq 0\], then
  \[P((A \cup B)|F) = P(A|F) + P(B|F) - P((A \cap B)|F)\]

• \[P(E'|F) = 1 - P(E|F)\]

• Conditional probability means probability under a given condition.

• The formula is \[P(A | B) = \frac{P(A \cap B)}{P(B)}\], where \[P(B) \neq 0\].

• First restrict the sample space according to the given condition, then count the favourable outcomes.

• The numerator represents outcomes common to both events.

• Do not confuse P(A | B) with P(B | A).

• For independent events, P(A | B) = P(A), provided P(B) ≠ 0.