Problem 63 - A First Course in Probability

Question

\(A\) and \(B\) are involved in a duel. The rules of the duel are that they are to pick up their guns and shoot at each other simultaneously. If one or both are hit, then the duel is over. If both shots miss, then they repeat the process. Suppose that the results of the shots are independent and that each shot of \(A\) will hit \(B\) with probability \(p_{A},\) and each shot of \(B\) will hit \(A\) with probability \(p_{B}\). What is (a) the probability that \(A\) is not hit? (b) the probability that both duelists are hit? (c) the probability that the duel ends after the \(n\) th round of shots? (d) the conditional probability that the duel ends after the \(n\) th round of shots given that \(A\) is not hit? (e) the conditional probability that the duel ends after the nth round of shots given that both duelists are hit?

Step-by-Step Solution

Answer

(a) The probability of A not being hit is \((1 - p_B)^n\), where n is the number of rounds. (b) The probability of both duelists being hit is \(\frac{p_A p_B}{1 - (1-p_A)(1-p_B)}\). (c) The probability of the duel ending after the nth round is \((1-p_A)^{n-1}(1-p_B)^{n-1}(p_A(1-p_B) + (1-p_A) p_B + p_A p_B)\). (d) The conditional probability of the duel ending after the nth round, given A is not hit, is \((1-p_A)^{n-1} p_B\). (e) The conditional probability of the duel ending after the nth round, given both duelists are hit, is \((1- p_A)^{n-1} (1-p_B)^{n-1} (1 - (1-p_A)(1-p_B))\).

Step 1: (a) Probability of A not being hit

First, let's find the probability of A not being hit by B. Given that B's shot hits A with probability \(p_B\), the probability of B missing A in one round is \(1 - p_B\). For A not to be hit, B must miss in every round. Since the shots in each round are independent, to find the probability of an event happening in all the rounds, we multiply the probabilities of each round. Thus, the probability of A not being hit is \((1 - p_B)^n\), where n is the number of rounds.

Step 2: (b) Probability of both duelists being hit

Now, we want to find the probability of both duelists being hit. This will only occur when both A and B hit their target in the same round. The probability of this happening in a single round is \(p_A * p_B\). To calculate the probability of this event happening across all rounds, we will sum the probability of it happening in each round multiplied by the probability of it not happening in all the previous rounds. So, for example, in round 2, the probability of it happening will be \((1-p_A)*(1-p_B)*p_A*p_B\) and so on. This is a geometric series. The probability of both duelists being hit is \(\sum_{n=1}^{\infty} (1-p_A)^{n-1} (1-p_B)^{n-1} p_A p_B = \frac{p_A p_B}{1 - (1-p_A)(1-p_B)}\).

Step 3: (c) Probability of the duel ending after the nth round

The probability of the duel ending after the nth round is the probability of one or both duelists getting hit only in the nth round, but not before that. For that to occur, both A and B must miss in the first (n-1) rounds. The probability of A and B missing their targets in the first (n-1) rounds is \((1-p_A)^{n-1}(1-p_B)^{n-1}\). Now, in the nth round, either A hits B, B hits A, or both hit each other. We can calculate the probability by summing up all the possible ways the duel can end: - A hits B, B misses A: \((1-p_A)^{n-1}(1-p_B)^{n-1} p_A (1-p_B)\) - A misses B, B hits A: \((1-p_A)^{n-1}(1-p_B)^{n-1} (1-p_A) p_B\) - A hits B, B hits A: \((1-p_A)^{n-1}(1-p_B)^{n-1} p_A p_B\) Hence, the probability of the duel ending after the nth round is: \((1-p_A)^{n-1}(1-p_B)^{n-1}(p_A(1-p_B) + (1-p_A) p_B + p_A p_B)\)

Step 4: (d) Conditional probability of the duel ending after the nth round, given A is not hit

Given that A is not hit, we need to find the conditional probability of the duel ending after the nth round. To calculate the conditional probability, we can use the formula: \(P(E | F) = \frac{P(E \cap F)}{P(F)}\) In this case, E is the event that the duel ends after the nth round, and F is the event that A is not hit. We have already calculated the probability of A not being hit and the probability of the duel ending after the nth round, so we can substitute those values into the formula. \(P(E | F) = \frac{(1-p_A)^{n-1} (1-p_B)^{n-1}(1-p_A)p_B}{(1-p_B)^n}\) This simplifies to: \(P(E | F) = (1-p_A)^{n-1} p_B\)

Step 5: (e) Conditional probability of the duel ending after the nth round, given both duelists are hit

Given that both duelists are hit, we want to find the conditional probability of the duel ending after the nth round. We can again use the formula: \(P(E | F) = \frac{P(E \cap F)}{P(F)}\) In this case, E is the event that the duel ends after the nth round, and F is the event that both duelists are hit. We have already calculated the probability of both duelists being hit and the probability of the duel ending after the nth round, so we can substitute those values into the formula. Here the intersection of E and F would be that the duel ends after the nth round due to both duelists being hit. The probability of this event is \((1-p_A)^{n-1} (1-p_B)^{n-1} p_A p_B\). \(P(E | F) = \frac{(1-p_A)^{n-1} (1-p_B)^{n-1} p_A p_B}{\frac{p_A p_B}{1 - (1-p_A)(1-p_B)}}\) This simplifies to: \(P(E | F) = (1- p_A)^{n-1} (1-p_B)^{n-1} (1 - (1-p_A)(1-p_B))\)

Key Concepts

Conditional Probability

Conditional probability helps us determine the likelihood of an event occurring, given that another event has already taken place. In the context of the duel scenario, imagine you want to know the chance of the duel ending specifically in the nth round, under the condition that a certain outcome (like one participant not being hit or both being hit) occurs.
To find the conditional probability of an event E occurring given event F, we use the formula:

• \( P(E | F) = \frac{P(E \cap F)}{P(F)} \)

Your understanding of how events are interlinked is crucial here. Always break the problem down by considering both the joint probability \( P(E \cap F) \) (the likelihood both events happen together) and the probability of the condition \( P(F) \). This can help simplify complex probability questions, much like solving a puzzle step by step.

Independent Events

Independent events refer to occurrences where the outcome of one does not affect the other. In the duel described, each shot is considered an independent event. This means that whether duelist A hits or misses B, it does not influence whether duelist B hits or misses A.
Mathematically, two events, say X and Y, are independent if the probability of both events occurring together is the product of the probabilities of each occurring alone:

• \( P(X \cap Y) = P(X) \times P(Y) \)

When events are independent, you can analyze and calculate probabilities for each event individually. This concept simplifies many calculations, as you can focus on each event separately without needing to consider the cascading effects on subsequent events.

Geometric Series

A geometric series is a sum of terms where each term is a fixed multiple of the previous one. In probability theory, it's commonly used to calculate the summation of probabilities that follow a specific repetitive pattern, like when events repeat until a certain condition is met.
In the duel, the scenario where rounds continue until one or both duelists hit involves a geometric series. The series formulates as:

• \( \sum_{n=1}^{\infty} r^{n-1} \cdot a = \frac{a}{1 - r} \)

where \( a \) is the first term with probability, and \( r \) is the common ratio. This helps in finding the total probability across infinite trials where an event may not immediately succeed, much like accumulating coins in a piggy bank till they're enough for a treat!

Probability Distribution

A probability distribution maps out every potential outcome of a random variable and assigns probabilities to these outcomes. In simpler terms, it is a comprehensive overview of where and how likely the various results might appear.
For the duel, the probability distribution describes the likelihood of different scenarios, such as duelists hitting or missing each other. Each outcome can be assigned a probability, forming a distribution. It's like a detailed prediction chart that tells you not only what can happen but how likely each situation is.

• If you have a proper understanding of probability distributions, you can anticipate the outcome landscape.
• This concept is fundamental in both theoretical scenarios like duels and real-world applications like weather forecasts or lottery predictions.

Understanding this allows you to visualize the entire possibility space, making sense of probabilities beyond just single events or rounds.