Construct a scenario where the independence of two events is violated, and demonstrate how this affects the computation of their combined probability.
user
Sure, here's an example to demonstrate this concept.

Let's consider the following two events:

1. Event A: Toss a coin.
2. Event B: Flip a coin.

In this scenario, Event A and Event B are not independent because the outcome of Event A (whether it is a head or tail) affects the outcome of Event B. If the outcome of Event A is a head, Event B is always a head, and if it's a tail, Event B is always a tail.

Now, let's assume that there are only two possible outcomes for each event, which are heads (H) or tails (T). So, the probability distribution for each event is binomial with one trial and two possible outcomes.

1. Calculate the probability distribution function for Event A (since this is a single trial, it's either heads or tails). The probability of getting heads is equal to the probability of getting tails, which is 0.5 (assuming a fair coin). Thus, Pr(A) = Pr(B) = Pr(H) = Pr(T) = 0.5.

2. Calculate the probability distribution function for Event B, using Bayes' theorem. Pr(H|A) = (Pr(A|H)·Pr(H)) / Pr(A) = (1·0.5) / 0.5 = 1.

Here, Pr(A|H) = 1 because if Event A occurs, it's a sure thing that Event B will be heads. Similarly, Pr(T|A) = 0 because if Event A occurs, then Event B cannot be tails.

Similarly, we have Pr(H|T) = 0 and Pr(T|T) = 1.

So, Pr(B) = Pr(H) * Pr(H|T) + Pr(T) * Pr(T|T) = 0.5 * 0 + 0.5 * 1 = 0.5

Hence, the probability of Flipping Head in Event B is 0.5.

However, if the tossing of the coin in Event A is done over successive trials, the probability that the outcome is Heads, say H1, H2, H3,...follows a geometric distribution.

This violates the assumption of independence in the sense that events A1, A2, A3,... Are not independent as the probability of A1 is not equal in each trial. The probability of getting Heads (or Tails) changes with each succeeding trial, even if the coin is back in its same starting position.

Therefore, the independence axiom in our coin-flipping case is violated. The probability, in contrast to the expected, changes across repeated As occurring in the geometric probability.

A. The probability of an event is usually constant, or distant characteristic that does not vary amongst trips, and it is identical in various editions (Alternatively, independent of the last) - in different words, Monotonous, so A always implies that pairs of occurrences cannot be independent.
assistant
Construct a scenario where the independence of two events is violated, and demonstrate how this affects the computation of their combined probability.
user
Scenario:

Consider a standard deck of 52 playing cards, where each card has one of four suits (hearts, diamonds, clubs, spades) and one of 13 ranks (Ace through 10, and the face cards Jack, Queen, King). Now, let's define two events:

Event A: Drawing a card that is a heart.
Event B: Drawing a card that is a Queen.

Under the assumption that the deck is well-shuffled, these two events are initially independent of each other. The probability of drawing a heart (Event A) is 13/52 (since there are 13 hearts in the deck), and the probability of drawing a Queen (Event B) is 4/52 (since there are 4 Queens in the deck).

To find the probability of both events happening together (drawing a heart and a Queen), we would typically multiply their individual probabilities, assuming they are independent: P(A and B) = P(A) * P(B) = (13/52) * (4/52) = 1/52.

Now, let's violate the independence by introducing a condition:

Condition: Remove the Queen of Hearts from the deck before drawing a card.

Now, the events are no longer independent since removing the Queen of Hearts affects the probability of both drawing a heart and drawing a Queen. With this condition, there are only 12 hearts and 3 Queens left in the deck.

Calculating the probabilities again, we get:
P(A) = Probability of drawing a heart = 12/51
P(B) = Probability of drawing a Queen = 3/51

However, if we try to calculate the probability of both events happening together using the formula for independent events, we get a incorrect result: P(A and B) = P(A) * P(B) = (12/51) * (3/51) = 36/2601, which is clearly not correct, as there is actually NO WAY to draw both a heart and a Queen from the deck after removing the Queen of Hearts.

The correct probability of drawing both a heart and a Queen with the given condition is 0, since they cannot occur together: P(A and B) = 0.

This demonstrates that when the independence of two events is violated, the formula for the combined probability of both events occurring (P(A and B) = P(A) * P(B)) is no longer valid, and may produce incorrect results.
assistant