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Chapter 30: Problem 5

Two duellists, \(A\) and \(B\), take alternate shots at each other, and the duel is over when a shot (fatal or otherwise!) hits its target. Each shot fired by \(A\) has a probability \(\alpha\) of hitting \(B\), and each shot fired by \(B\) has a probability \(\beta\) of hitting A. Calculate the probabilities \(P_{1}\) and \(P_{2}\), defined as follows, that \(A\) will win such a duel: \(P_{1}, A\) fires the first shot; \(P_{2}, B\) fires the first shot. If they agree to fire simultaneously, rather than alternately, what is the probability \(P_{3}\) that \(A\) will win, i.e. hit \(B\) without being hit himself?

Short Answer

Expert verified
The probabilities are: \(P_{A} = \frac{\alpha}{\alpha + \beta - \alpha\beta}\), \(P_{B} = (1-\beta) \frac{\alpha}{\alpha + \beta - \alpha\beta}\) and \(P_{3} = \alpha(1-\beta)\).

Step by step solution

01

Define the Winning Probabilities

We need to calculate the probabilities that \(A\) will win under different scenarios. Let's denote \(P_{A}\) as the probability of \(A\) winning when \(A\) fires first and \(P_{B}\) as the probability of \(A\) winning when \(B\) fires first.
02

Calculate \(P_{A}\)

If \(A\) fires first, the duel is over when the first hit occurs.\(A\)'s probability of hitting \(B\) on the first shot is \(\alpha\). If \(A\) misses, \(B\) fires next. The probability of \(A\) winning after the first miss is the probability of \(B\) missing \((1-\beta)\) times the probability of \(A\) winning if it was \(A\)'s turn again. More formally, the equation is: \[ P_{A} = \alpha + (1-\alpha)(1-\beta)P_{A} \]
03

Solve the Equation for \(P_{A}\)

We now solve for \(P_{A}\): \[ P_{A} = \alpha + (1-\alpha)(1-\beta)P_{A} \] Factor out \(P_{A}\) from the right: \[ P_{A} = \alpha + P_{A}(1 - \alpha - \beta + \alpha\beta) \] Isolate \(P_{A}\): \[ P_{A}(1 - (1 - \alpha - \beta + \alpha\beta)) = \alpha \] Simplify: \[ P_{A}(\alpha + \beta - \alpha\beta) = \alpha \] Thus we have: \[ P_{A} = \frac{\alpha}{\alpha + \beta - \alpha\beta} \]
04

Calculate \(P_{B}\)

If \(B\) fires first, the duel's outcome equations follow a similar pattern. Using the same reasoning as before, we have: \[ P_{B} = (1-\beta)P_{A} \] Substitute \(P_{A}\) from Step 3 into this equation: \[ P_{B} = (1-\beta) \frac{\alpha}{\alpha + \beta - \alpha\beta} \]
05

Calculate \(P_{3}\) - Simultaneous Fire

When both fire simultaneously, \(A\) will win if \(A\) hits \(B\) and \(B\) misses \(A\). Therefore, the probability of this event is given by: \[ P_{3} = \alpha(1-\beta) \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

conditional probability
Conditional probability, represented as \(P(A|B)\), is the probability of event \(A\) occurring given that event \(B\) has occurred. This concept is fundamental in the analysis of the duel problem because the outcome of each shot can affect the subsequent probabilities. In the duel scenario:
  • If \(A\) misses, the conditional probability of winning changes based on whether \(B\) hits or misses.
  • The calculations incorporate these changing conditions, thus introducing the concept of conditional probability.
For instance, if \(A\) shoots first and misses, the probability of \(A\) winning then depends on \(B\)'s next shot. If \(B\) misses, the condition changes again and \(A\)'s turn becomes tantamount again. Mathematically, we derive:
\[ P_{A} = \alpha + (1-\alpha )(1-\beta)P_{A} \] to encapsulate the probability of \(A\) hitting right away or both missing with \(A\) having another turn.
duel problem
The duel problem is a classic example in probability theory. Here, two individuals engage in a duel with alternating shots that continue until one hits the other. The aim is to calculate the winning probabilities.
  • Duel problems introduce real-world scenarios where probabilities are not just static numbers but depend on sequences of events.
  • They help in understanding complex probabilistic systems by breaking them into simpler sequential and conditional events.
Consider our example, with duelists \(A\) and \(B\):
  • \(A\) starts, and his hitting probability is \(\alpha\).
  • If \(A\) misses, \(B\) gets his chance with a hitting probability \(\beta\).
  • The developed form of probabilities reveals the interplay of circumstances and time over repeated rounds.
This setup leads us to the equations from which we solved for the probability of \(A\) winning, \(P_{A}\). Similarly, we determined \(P_{B}\) when \(B\) fires first.
sequential probability
Sequential probability involves the analysis of outcomes that occur in a specific sequence. Each shot in the duel is a sequential event with its probability affected by the previous shot. This concept is crucial in the duel problem:
  • First, we determine the probability of a single event (a shot hitting or missing).
  • Then, these probabilities are multiplied as per order of events, forming a sequence.
For instance:
  • When \(A\) fires first, the probability explanation and subsequent outcome depend on \(\alpha\), with the sequence followed by \(\beta\).
  • In the alternate turn scenario, sequential probability represents how missing shots affect who gets the next turn.
Ultimately, solving for \(P_{A}\) and \(P_{B}\) makes these dependencies clear. We factor in each sequential step:
\[ P_{A} = \alpha + P_{A}(1- \alpha- \beta +\alpha \beta) \], ensuring accurate representation of conditional and sequential events.

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Most popular questions from this chapter

As assistant to a celebrated and imperious newspaper proprietor, you are given the job of running a lottery, in which each of his five million readers will have an equal independent chance, \(p\), of winning a million pounds; you have the job of choosing \(p\). However, if nobody wins it will be bad for publicity, whilst if more than two readers do so, the prize cost will more than offset the profit from extra circulation - in either case you will be sacked! Show that, however you choose \(p\), there is more than a \(40 \%\) chance you will soon be clearing your desk.

An electronics assembly firm buys its microchips from three different suppliers; half of them are bought from firm \(X\), whilst firms \(Y\) and \(Z\) supply \(30 \%\) and \(20 \%\), respectively. The suppliers use different quality- control procedures and the percentages of defective chips are \(2 \%, 4 \%\) and \(4 \%\) for \(X, Y\) and \(Z\), respectively. The probabilities that a defective chip will fail two or more assembly-line tests are \(40 \%, 60 \%\) and \(80 \%\), respectively, whilst all defective chips have a \(10 \%\) chance of escaping detection. An assembler finds a chip that fails only one test. What is the probability that it came from supplier \(X\) ?

A continuous random variable \(X\) is uniformly distributed over the interval \([-c, c]\). A sample of \(2 n+1\) values of \(X\) is selected at random and the random variable \(Z\) is defined as the median of that sample. Show that \(Z\) is distributed over \([-c, c]\) with probability density function $$ f_{n}(z)=\frac{(2 n+1) !}{(n !)^{2}(2 c)^{2 n+1}}\left(c^{2}-z^{2}\right)^{n} $$ Find the variance of \(Z\).

A continuous random variable \(X\) has a probability density function \(f(x) ;\) the corresponding cumulative probability function is \(F(x)\). Show that the random variable \(Y=F(X)\) is uniformly distributed between 0 and 1 .

A husband and wife decide that their family will be complete when it includes two boys and two girls - but that this would then be enough! The probability that a new baby will be a girl is \(p\). Ignoring the possibility of identical twins, show that the expected size of their family is $$ 2\left(\frac{1}{p q}-1-p q\right) $$ where \(q=1-p\)
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