Conditional Probability

Introduction to Conditional Probability

In many situations, we are interested in the probability of an event happening, given that we already have some additional information. For instance, what is the probability that it will rain today, given that the morning is cloudy? This 'given' information restricts our focus.

Definition: If $A$ and $B$ are two events associated with the same sample space $S$ of a random experiment, then the probability of occurrence of event $A$ under the condition that $B$ has already occurred (and $P(B) \neq 0$) is called the conditional probability of $A$ given $B$. It is denoted by $P(A|B)$.

The Formula: $$P(A|B) = \frac{P(A \cap B)}{P(B)}, \quad P(B) \neq 0$$

Similarly, the conditional probability of $B$ given $A$ is: $$P(B|A) = \frac{P(A \cap B)}{P(A)}, \quad P(A) \neq 0$$

Intuition: Imagine the sample space $S$ shrinks to just the event $B$. We are now looking for the portion of $A$ that lies within this new sample space $B$.

Properties of Conditional Probability

Let $E$ and $F$ be events of a sample space $S$ and $G$ be another event such that $P(G) \neq 0$. The following properties hold:

• Property 1: $P(S|G) = 1$ and $P(G|G) = 1$.
• Property 2: 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: $$P((A \cup B)|F) = P(A|F) + P(B|F) - P((A \cap B)|F)$$ Note: If $A$ and $B$ are disjoint, then $$P((A \cup B)|F) = P(A|F) + P(B|F)$$
• Property 3: $$P(E'|F) = 1 - P(E|F)$$ This is the complement rule for conditional probability.

Example 1: Basic Coin Toss

A coin is tossed three times. Find the probability that at least two heads appear, given that at least one head appeared.

Solution: Step 1: Write the sample space $S$. $$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$$ Hence, $$n(S) = 8$$

Step 2: Define the events. Let $A$ = event of getting at least two heads. $$A = \{HHH, HHT, HTH, THH\}$$ So, $$n(A) = 4$$ Let $B$ = event of getting at least one head. $$B = \{HHH, HHT, HTH, THH, HTT, THT, TTH\}$$ So, $$n(B) = 7$$

Step 3: Find $A \cap B$. Since every outcome of $A$ already belongs to $B$, $$A \cap B = A$$ Therefore, $$n(A \cap B) = 4$$

Step 4: Apply the conditional probability formula. $$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{4/8}{7/8} = \frac{4}{7}$$

Answer: $$\boxed{\frac{4}{7}}$$

Example 2: Dice and Conditions

A black and a red die are rolled. Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.

Solution: Step 1: The total number of outcomes is $$6 \times 6 = 36$$

Step 2: Let $B$ be the event that the black die shows 5. $$B = \{(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)\}$$ Hence, $$n(B) = 6 \quad \Rightarrow \quad P(B) = \frac{6}{36}$$

Step 3: Let $A$ be the event that the sum is greater than 9. Possible sums are 10, 11, and 12, so $$A = \{(4,6), (5,5), (6,4), (5,6), (6,5), (6,6)\}$$

Step 4: Find $A \cap B$. We need outcomes where the black die is 5 and the sum is greater than 9: $$A \cap B = \{(5,5), (5,6)\}$$ Thus, $$n(A \cap B) = 2 \quad \Rightarrow \quad P(A \cap B) = \frac{2}{36}$$

Step 5: Compute the conditional probability. $$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{2/36}{6/36} = \frac{2}{6} = \frac{1}{3}$$

Answer: $$\boxed{\frac{1}{3}}$$

Example 3: Families with Children

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

Solution: Step 1: Assuming each child is equally likely to be a boy or a girl, the sample space is $$S = \{BB, BG, GB, GG\}$$ Thus, $$n(S) = 4$$

Step 2: Define the events. Let $A$ = event that both children are boys. $$A = \{BB\}$$ Let $B$ = event that at least one child is a boy. $$B = \{BB, BG, GB\}$$

Step 3: Find the intersection. $$A \cap B = \{BB\}$$ So, $$P(A \cap B) = \frac{1}{4}, \qquad P(B) = \frac{3}{4}$$

Step 4: Use the formula. $$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{1/4}{3/4} = \frac{1}{3}$$

Answer: $$\boxed{\frac{1}{3}}$$

Example 4: Cards Selection

Ten cards numbered 1 to 10 are placed in a box. One card is drawn. If it is known that the number on the drawn card is more than 3, what is the probability that it is an even number?

Solution: Step 1: The sample space is $$S = \{1,2,3,4,5,6,7,8,9,10\}$$

Step 2: Let $B$ be the event that the number is more than 3. $$B = \{4,5,6,7,8,9,10\}$$ Hence, $$n(B) = 7$$

Step 3: Let $A$ be the event that the number is even. $$A = \{2,4,6,8,10\}$$

Step 4: Find $A \cap B$. $$A \cap B = \{4,6,8,10\}$$ Hence, $$n(A \cap B) = 4$$

Step 5: Therefore, $$P(A|B) = \frac{n(A \cap B)}{n(B)} = \frac{4}{7}$$

Answer: $$\boxed{\frac{4}{7}}$$

Example 5: Using $P(A)$, $P(B)$, and $P(A \cup B)$

Given $P(A) = 0.6, P(B) = 0.3$ and $P(A \cup B) = 0.7$, find $P(A|B)$.

Solution: Step 1: Use the addition formula: $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ Substitute the values: $$0.7 = 0.6 + 0.3 - P(A \cap B)$$ $$0.7 = 0.9 - P(A \cap B)$$ $$P(A \cap B) = 0.2$$

Step 2: Now use the conditional probability formula: $$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.2}{0.3} = \frac{2}{3}$$

Answer: $$\boxed{\frac{2}{3}}$$

Example 6: Complement Rule

If $P(A) = 0.8, P(B) = 0.5$ and $P(B|A) = 0.4$, find $P(A'|B)$.

Solution: Step 1: Use the definition of conditional probability: $$P(B|A) = \frac{P(A \cap B)}{P(A)}$$ Substitute the values: $$0.4 = \frac{P(A \cap B)}{0.8}$$ Hence, $$P(A \cap B) = 0.4 \times 0.8 = 0.32$$

Step 2: Find $P(A|B)$. $$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.32}{0.5} = 0.64$$

Step 3: Use the complement rule. $$P(A'|B) = 1 - P(A|B) = 1 - 0.64 = 0.36$$

Answer: $$\boxed{0.36}$$

Example 7: School Statistics

In a school, there are 1000 students, out of which 430 are girls. It is known that out of 430, 10% of the girls study in class XII. What is the probability that a student chosen randomly studies in class XII given that the chosen student is a girl?

Solution: Step 1: Let

• $G$ = event that the selected student is a girl,
• $E$ = event that the selected student studies in class XII.

Step 2: Number of girls is $$n(G) = 430$$ Out of these, 10% study in class XII. Therefore, $$n(E \cap G) = 10\% \text{ of } 430 = 43$$

Step 3: Now, $$P(E|G) = \frac{n(E \cap G)}{n(G)} = \frac{43}{430} = 0.1$$

Answer: $$\boxed{0.1}$$

Example 8: Two Dice Sum

A die is thrown twice and the sum of the numbers appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once?

Solution: Step 1: Let $B$ be the event that the sum is 6. Possible outcomes are: $$B = \{(1,5), (2,4), (3,3), (4,2), (5,1)\}$$ Hence, $$n(B) = 5$$

Step 2: Let $A$ be the event that the number 4 appears at least once.

Step 3: Find $A \cap B$. Among the outcomes of $B$, the ones containing 4 are: $$A \cap B = \{(2,4), (4,2)\}$$ Thus, $$n(A \cap B) = 2$$

Step 4: Therefore, $$P(A|B) = \frac{n(A \cap B)}{n(B)} = \frac{2}{5}$$

Answer: $$\boxed{\frac{2}{5}}$$

Example 9: Property $P((A \cup B)|F)$

Given $P(A|F) = 0.3, P(B|F) = 0.2$ and $P((A \cap B)|F) = 0.1$, find $P((A \cup B)|F)$.

Solution: Step 1: Use the formula $$P((A \cup B)|F) = P(A|F) + P(B|F) - P((A \cap B)|F)$$

Step 2: Substitute the values: $$P((A \cup B)|F) = 0.3 + 0.2 - 0.1 = 0.4$$

Answer: $$\boxed{0.4}$$

Example 10: Logic Puzzle

Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that the youngest is a girl?

Solution: Step 1: Write the sample space in the order (youngest, oldest): $$S = \{(B,B), (B,G), (G,B), (G,G)\}$$

Step 2: Let

• $A$ = event that both are girls,
• $B$ = event that the youngest is a girl. Then, $$A = \{(G,G)\}$$ $$B = \{(G,B), (G,G)\}$$ Hence, $$n(B) = 2$$

Step 3: Find the intersection: $$A \cap B = \{(G,G)\}$$ So, $$n(A \cap B) = 1$$

Step 4: Therefore, $$P(A|B) = \frac{1}{2}$$

Answer: $$\boxed{\frac{1}{2}}$$