Question

Each of four persons fires one shot at a target. Let \(C_{k}\) denote the event that the target is hit by person \(k, k=1,2,3,4\). If \(C_{1}, C_{2}, C_{3}, C_{4}\) are independent and if \(P\left(C_{1}\right)=P\left(C_{2}\right)=0.7, P\left(C_{3}\right)=0.9\), and \(P\left(C_{4}\right)=0.4\), compute the probability that (a) all of them hit the target; (b) exactly one hits the target; (c) no one hits the target; (d) at least one hits the target.

Short answer

The probability that (a) all of them hit the target is \(0.7 * 0.7 * 0.9 * 0.4\). The probability that (b) exactly one hits the target is \(P(\(C_{1}\)) * P(\~\(C_{2}\)) * P(\~\(C_{3}\)) * P(\~\(C_{4}\)) + P(\~\(C_{1}\)) * P(\(C_{2}\)) * P(\~\(C_{3}\)) * P(\~\(C_{4}\)) + P(\~\(C_{1}\)) * P(\~\(C_{2}\)) * P(\(C_{3}\)) * P(\~\(C_{4}\)) + P(\~\(C_{1}\)) * P(\~\(C_{2}\)) * P(\~\(C_{3}\)) * P(\(C_{4}\)). The probability that (c) no one hits the target is \(P(\~\(C_{1}\)) * P(\~\(C_{2}\)) * P(\~\(C_{3}\)) * P(\~\(C_{4}\)\). The probability that (d) at least one hits the target is \(1 - P(\~\(C_{1}\)) * P(\~\(C_{2}\)) * P(\~\(C_{3}\)) * P(\~\(C_{4}\)\).

Step-by-step solution

Determine the probability that all of them hit the target

Since the events are independent, the probability that they all hit the target is the product of their individual hit probabilities. Hence, \(P(\(C_{1}\) \cap \(C_{2}\) \cap \(C_{3}\) \cap \(C_{4}\)) = P(\(C_{1}\)) * P(\(C_{2}\)) * P(\(C_{3}\)) * P(\(C_{4}\)) = 0.7 * 0.7 * 0.9 * 0.4.

Determine the probability that exactly one hits the target

There are four scenarios here: either of the four individuals being the one who hits the target. For each individual, this probability is the product of their hit probability and the miss probabilities of the other three. These four probabilities have to be added together. Hence, \(P(\(C_{1}\)) * P(\~\(C_{2}\)) * P(\~\(C_{3}\)) * P(\~\(C_{4}\)) + P(\~\(C_{1}\)) * P(\(C_{2}\)) * P(\~\(C_{3}\)) * P(\~\(C_{4}\)) + P(\~\(C_{1}\)) * P(\~\(C_{2}\)) * P(\(C_{3}\)) * P(\~\(C_{4}\)) + P(\~\(C_{1}\)) * P(\~\(C_{2}\)) * P(\~\(C_{3}\)) * P(\(C_{4}\)). Remember that the probability of not hitting is 1 minus the probability of hitting.

Determine the probability that no one hits the target

This is the product of the miss probabilities of all four individuals. Hence, \(P(\~\(C_{1}\)) * P(\~\(C_{2}\)) * P(\~\(C_{3}\)) * P(\~\(C_{4}\)).

Determine the probability that at least one hits the target

This can be determined as the complement of the event that no one hits the target. Hence, 1 - \(P(\~\(C_{1}\)) * P(\~\(C_{2}\)) * P(\~\(C_{3}\)) * P(\~\(C_{4}\)).

Key concepts

Independent Events in Probability

Understanding the notion of independent events is crucial in probability theory. Two or more events are considered independent if the occurrence of one does not affect the occurrence of the others. In our exercise, each person firing at the target represents an independent event, meaning person 1 hitting the target does not influence whether person 2, 3, or 4 will hit or miss the target. When calculating the probability of independent events occurring together, we simply multiply their individual probabilities.

For example, to find the probability that all persons hit the target, you multiply the probability of each person hitting the target. This fundamental property of independent events allows for straightforward calculations, which is particularly powerful when examining scenarios with multiple actors or components, like the firing exercise.

Probability Theory Fundamentals

At the heart of our exercise lies probability theory, a branch of mathematics concerned with the analysis of random phenomena. The core principle is to assign a numerical value, known as probability, to the likelihood of an event occurring; this value ranges from 0 (impossibility) to 1 (certainty). Probability calculations require a clear understanding of events and outcomes.

When dealing with multiple events, as in our example with multiple shooters, probability theory provides rules and formulas to calculate the likelihood of various outcomes, such as 'exactly one hits the target' or 'at least one hits the target'. It's essential to apply the correct methods - like addition for mutually exclusive events or multiplication for independent events - to achieve accurate results.

Using the Complement Rule

The complement rule is a powerful shortcut in probability that simplifies calculations, particularly for complex problems. The rule states that the probability of an event not happening is equal to one minus the probability of the event happening. Formally, if 'A' is an event, then the probability of 'not A' is given by: \( P(\sim A) = 1 - P(A) \).

In the context of our exercise, the complement rule is used to calculate the probability that at least one person hits the target (event 'A') by subtracting the probability that no one hits the target (event '\(\sim A\)') from 1. This is much simpler than adding up all the various probabilities of one or more persons hitting the target. By mastering the complement rule, students can tackle more challenging probability questions with greater ease and accuracy.