Question

An Olympic archer is able to hit the bull's-eye \(80 \%\) of the time. Assume each shot is independent of the others. If she shoots 6 arrows, what's the probability of each of the following results? a) Her first bull's-eye comes on the third arrow. b) She misses the bull's-eye at least once. c) Her first bull's-eye comes on the fourth or fifth arrow. d) She gets exactly 4 bull's-eyes. e) She gets at least 4 bull's-eyes. f) She gets at most 4 bull's-eyes.

Short answer

a) 0.032; b) 0.737856; c) 0.00768; d) 0.24576; e) 0.90112; f) 0.34464.

Step-by-step solution

Clarify the Variables

Denote the probability of hitting a bull's-eye as \( p = 0.8 \). Each shot is independent, which means a binomial distribution with \( n = 6 \) and \( p = 0.8 \) can be applied for multiple shots.

Calculate (a) The Probability That the First Bull's-eye Comes on the Third Arrow

For the first bull's-eye to come on the third arrow, the archer must miss the first two and hit the third: \[P( ext{first bull's-eye on third}) = (1-p)(1-p)p = (0.2)(0.2)(0.8) = 0.032.\]

Calculate (b) The Probability of Missing the Bull's-eye at Least Once

Calculate the probability of hitting all six bull's-eyes first: \[P( ext{hitting all}) = p^6 = 0.8^6 = 0.262144.\]Then find the complement: \[P( ext{miss at least once}) = 1 - P( ext{hitting all}) = 1 - 0.262144 = 0.737856.\]

Calculate (c) The Probability That the First Bull's-eye Comes on the Fourth or Fifth Arrow

Calculate for the fourth arrow: \[P( ext{first on fourth}) = (1-p)^3p = (0.2)^3(0.8) = 0.0064.\]Calculate for the fifth arrow: \[P( ext{first on fifth}) = (0.2)^4(0.8) = 0.00128.\]Add both probabilities: \[P( ext{fourth or fifth}) = 0.0064 + 0.00128 = 0.00768.\]

Calculate (d) The Probability of Getting Exactly 4 Bull's-eyes

Use the binomial formula: \[P(X=4) = \binom{6}{4} p^4 (1-p)^2 = 15 \cdot (0.8)^4 (0.2)^2 = 0.24576.\]

Calculate (e) The Probability of Getting at Least 4 Bull's-eyes

Calculate the probabilities for 4, 5, and 6 bull's-eyes and sum them.- For 5 bull's-eyes: \[P(X=5) = \binom{6}{5} p^5 (1-p)^1 = 6 \cdot (0.8)^5 (0.2) = 0.393216.\]- For 6 bull's-eyes: \[P(X=6) = \binom{6}{6} p^6 (1-p)^0 = 1 \cdot (0.8)^6 = 0.262144.\]Sum probabilities for 4, 5, and 6: \[P(X\geq4) = 0.24576 + 0.393216 + 0.262144 = 0.90112.\]

Calculate (f) The Probability of Getting at Most 4 Bull's-eyes

This is the complement of getting more than 4 bull's-eyes: \[P(X\leq4) = 1 - P(X\geq5).\]Calculate for at least 5 bull's-eyes: \[P(X\geq5) = 0.393216 + 0.262144 = 0.65536.\]Then, \[P(X\leq4) = 1 - 0.65536 = 0.34464.\]

Key concepts

Probability Theory

Probability theory is a fundamental branch of mathematics that deals with predicting the likelihood of various outcomes. It's especially useful for understanding random processes and making informed decisions based on numerical data.

In probability theory, each potential outcome of a random experiment is assigned a likelihood, which is a number between 0 and 1. A probability of 0 means the event cannot happen, while a probability of 1 means it is certain to occur.

For example, when considering an event like an Olympic archer hitting a bull's-eye 80% of the time, probability theory allows us to calculate the likelihood of different sequences of hits and misses as demonstrated in the provided exercise.

• It helps practitioners calculate probabilities using principles like permutation and combination.
• The probabilities for independent events are multiplied together.
• These calculations give insight into the likelihood of various combined events, such as missing a target at least once out of several tries.

Overall, probability theory provides the tools to solve real-world problems, making it essential in fields ranging from sports to finance.

Independent Events

Independent events in probability theory are those whose outcomes do not affect each other. When one event occurs, it does not change the probability of the other event occurring.

In our example with the archer, each arrow shot is an independent event. That means whether the archer hits or misses the bull's-eye with one arrow does not influence the chance of hitting the bull's-eye with another arrow.

• This independence allows us to calculate the probability of a series of events by simply multiplying the probabilities of each individual event.
• In the exercise, calculating the chance of missing the first two arrows and hitting the third for the first bull's-eye involves multiplying the probabilities of missing and hitting separately.

This concept is crucial when dealing with problems where multiple outcomes occur in sequence and each outcome is separate, as it simplifies the process of determining overall probabilities.

Combinatorics

Combinatorics is the field of mathematics primarily concerned with counting, arrangement, and combination of objects. It plays a key role in calculating probabilities, especially when dealing with binomial distributions.

In the context of the given exercise, combinatorics helps us calculate the different ways to arrange bull's-eyes and misses across multiple attempts. Specifically, we can use combinations to determine how many ways archer can achieve a specified number of bull's-eyes out of a fixed number of shots.

• A common technique is using the binomial coefficient, represented as \(\binom{n}{k}\), where \(n\) is the total number of events, and \(k\) is the number of successful events.
• This coefficient tells us how many ways \(k\) successes can occur in \(n\) trials.
• It's integral to the binomial probability formula used to find probabilities for exact numbers of successes, like exactly four bull's-eyes.

Combinatorics transforms complex counting problems into manageable calculations, making it invaluable in probability distribution analysis.

Probability Distribution

A probability distribution describes how probabilities are distributed over the values of a random variable. For discrete variables, like those in our example, a probability distribution can explain the likelihood of each possible outcome.

In the exercise, we are dealing with a binomial distribution—a type of discrete probability distribution representing the number of successes in a fixed number of independent trials.

• Each trial, or arrow shot, results in one of two possible outcomes: hitting the bull's-eye or missing it.
• The binomial distribution is characterized by two parameters: the number of trials \(n\) and the probability of success in a single trial \(p\).

Key properties of a probability distribution include:

• The sum of all probabilities equals 1, ensuring that one of the possible outcomes will occur.
• Each probability value must lie between 0 and 1.

This distribution helps us understand probabilities for different numbers of successes, such as getting at least four bull's-eyes, a crucial outcome in the problem provided.