Conditional probability measures the probability of an event occurring given that another event has already occurred. It answers questions like “What’s the probability of rain given that it’s cloudy?” This concept is crucial in fields like statistics, machine learning, and decision-making.
Definition: The conditional probability of event A given event B, denoted as P(A|B), is the probability that A occurs, assuming B has already happened.
Here, P(A ∩ B) is the joint probability of A and B, and P(B) is the marginal probability of B.
Joint probability refers to the probability of two or more events occurring simultaneously. For events A and B, it’s denoted as P(A ∩ B) or P(A and B).
Formula: If A and B are independent, P(A ∩ B) = P(A) × P(B). For dependent events, it’s P(A) × P(B|A).
Joint probability is essential for understanding how events co-occur and forms the basis for conditional probability calculations.
Marginal probability is the probability of a single event occurring, regardless of other events. It’s obtained by summing (or integrating) joint probabilities over the other variables.
Formula: For discrete events, P(A) = Σ P(A ∩ B_i), where B_i are mutually exclusive and exhaustive events.
Marginal probabilities are the “totals” in probability tables and help in computing conditional probabilities.
Basic Formula: P(A|B) = P(A ∩ B) / P(B)
Multiplication Rule: P(A ∩ B) = P(A|B) × P(B) = P(B|A) × P(A)
Chain Rule: For multiple events, P(A ∩ B ∩ C) = P(A) × P(B|A) × P(C|A ∩ B)
Properties:
These properties allow us to update probabilities based on new information and solve complex problems.
If two events A and B are independent, the occurrence of one does not affect the probability of the other. In this case:
And P(A ∩ B) = P(A) × P(B). This simplifies calculations when events don’t influence each other, like flipping two coins.
To solidify your understanding, here are 6 quality practice questions with detailed answers. These cover various scenarios, including the boy-girl paradox variations and more.
Question 1: A family has two children. What is the probability that both are boys?
Answer:
Using marginal probability approach:
Sample space (S) = {(GG), (BG), (GB), (BB)} = 4 outcomes (assuming equal likelihood).
Favorable outcome = {(BB)} = 1
Required probability = 1/4
Question 2: A family has two children. What is the probability that both are boys given that at least one is a boy?
Answer:
Reduced sample space (given at least one boy) = {(BG), (GB), (BB)} = 3
Favorable outcome = {(BB)} = 1
Required probability = 1/3
This is P(Both boys | At least one boy) = P(Both boys ∩ At least one boy) / P(At least one boy) = (1/4) / (3/4) = 1/3
Question 3: A family has two children. What is the probability that both are boys given that the elder child is a boy?
Answer:
Reduced sample space (given elder is boy) = {(BG), (BB)} = 2
Favorable outcome = {(BB)} = 1
Required probability = 1/2
Here, the condition specifies the elder child, making it different from the previous case.
Question 4: A fair coin is flipped twice. What is the probability of getting heads on the second flip given that the first flip was tails?
Answer:
Since coin flips are independent, P(Second heads | First tails) = P(Second heads) = 1/2
Sample space: {(HH), (HT), (TH), (TT)}
Given first tails: Reduced space = {(TH), (TT)}
Favorable (second heads): {(TH)} = 1 out of 2, so 1/2
This illustrates that for independent events, conditional probability equals marginal probability.
Question 5: In a deck of 52 cards, what is the probability of drawing a king given that the card is a face card?
Answer:
Face cards: 12 (4 kings, 4 queens, 4 jacks)
P(King | Face card) = Number of kings / Number of face cards = 4/12 = 1/3
Using formula: P(King ∩ Face card) / P(Face card) = (4/52) / (12/52) = 4/12 = 1/3
This is a dependent scenario as the condition reduces the sample space.
Question 6: The probability of rain is 0.3. If it rains, the probability of traffic is 0.8; if not, it’s 0.2. What is the probability of traffic?
Answer:
Using total probability theorem:
P(Traffic) = P(Traffic | Rain) × P(Rain) + P(Traffic | No Rain) × P(No Rain)
= (0.8 × 0.3) + (0.2 × 0.7) = 0.24 + 0.14 = 0.38
This demonstrates marginal probability as a sum of conditional probabilities weighted by priors.
Question 7: A bag contains 5 red and 3 blue marbles. Two marbles are drawn without replacement. What is the probability that the second marble is red given that the first is red?
Answer:
This is a dependent events scenario because drawing the first marble affects the sample space for the second draw.
Total marbles: 5 red + 3 blue = 8
P(First red) = 5/8
Given first marble is red, remaining marbles: 4 red + 3 blue = 7
P(Second red | First red) = Number of remaining red / Remaining total = 4/7
Using formula: P(Second red ∩ First red) / P(First red) = [(5/8) × (4/7)] / (5/8) = 4/7
Thus, the probability that the second marble is red given the first is red is 4/7.
These questions cover a range of scenarios to build your intuition. Practice more to master conditional probability!
