91Ó°ÊÓ

Whenever we talk about independent events, we're referring to situations where the outcome of one event does not affect the outcome of another. When tossing a coin, each toss is independent of the others. This means the probability of getting a head or a tail remains constant at each toss, regardless of past outcomes.

In mathematical terms, two events, say \(A\) and \(B\), are independent if the probability of both occurring together is simply the product of their individual probabilities. Formally, this is written as \( \mathbb{P}(A \cap B) = \mathbb{P}(A) \times \mathbb{P}(B) \).

• This concept is crucial when analyzing sequences of events, as shown in our example where coin tosses are independent, each with a head probability \(p\) and tail probability \(1-p\).
• If the outcomes were not independent, calculating the probabilities of specific sequence patterns, like consecutive heads or tails, would require a different approach.

Understanding this concept of independence allows us to simplify complex problems by considering each event separately, provided they adhere to this independent nature.

The probability of sequences involves calculating the chance of a specific order of events occurring. In the context of our problem, we're interested in sequences of successive heads or tails, such as seeing \(r\) consecutive heads before \(s\) consecutive tails.

To tackle such problems, we consider the probabilities of each outcome and the independence of events. For successive heads, if the first toss is a head, the chance of seeing an additional \(r-1\) heads is \( p^{r-1} \).

• This exponential term \( p^{r-1} \) arises because each head is an independent event with probability \( p \), and the occurrences multiply together.
• Similarly, sequences of tails are calculated using \( (1-p)^{s-1} \) for seeing \(s-1\) additional tails after the first tail.

These probabilities help us determine whether a sequence of events, like the first run of heads or tails, occurs first.
By employing these calculations, we can find how likely it is for certain sequences to happen in any stochastic process of independent events.

The Law of Total Probability is a fundamental concept that helps in finding the probability of an event by considering all possible ways that event can occur. In our exercise, we need this law to determine the overall probability \( \mathbb{P}(E) \) of the first \(r\) heads occurring before \(s\) tails.

To apply it, we consider all possibilities based on the outcome of the first toss. This breaks down into:

• \( \mathbb{P}(E \mid A=\text{head}) \): The probability of event \(E\) given the first toss is a head.
• \( \mathbb{P}(E \mid A=\text{tail}) \): The probability of event \(E\) given the first toss is a tail.

We

Working with a system of equations is a powerful strategy when dealing with probabilities involving multiple conditions. In this problem, we are given equations \( \mathbb{P}(E \mid A=\text{head}) \) and \( \mathbb{P}(E \mid A=\text{tail}) \). These equations form a system that we need to solve to find individual probabilities.

The trick is to express each conditional probability in terms of the other and solve for one, allowing us to find the other directly using substitution.

• First, assume \( \mathbb{P}(E \mid A=\text{head}) = y \) and \( \mathbb{P}(E \mid A=\text{tail}) = x \).
• Substitute \( y \) into the equation for \( x \) or vice-versa. This permits one equation to contain only one variable, simplifying the problem.
• This mathematical maneuvering results in explicit values for both \( x \) and \( y \), the probabilities of event \(E\) given different starting points.

This process illustrates how resolving a system of equations is essential in deducing unknown probabilities when faced with multiple pathways to the same outcome.