Question

The probability of hitting a target on a shot is \((2 / 3)\). If a person fires 8 shots at a target, Let \(X\) denote the number of times he hits the target, and find: (a) \(\mathrm{P}(\mathrm{X}=3)\) (b) \(\mathrm{P}(1<\mathrm{X} \leq 6)\) (c) \(\mathrm{P}(\mathrm{X}>3)\).

Short answer

The probability for each scenario is as follows: (a) $\mathrm{P}(X=3) \approx 0.129$ (b) $\mathrm{P}(1< X \leq 6) \approx 0.346$ (c) $\mathrm{P}(X>3) \approx 0.851$.

Step-by-step solution

Parameters and Definitions

Let's first define the parameters for our problem: - Number of trials (shots), \(n = 8\) - Number of successes (hits), \(X\) - Probability of success on each trial, \(p = \frac{2}{3}\) - Probability of failure on each trial, \(1-p = \frac{1}{3}\) Now, we will calculate the probabilities for scenarios (a), (b), and (c).

(a) \(\mathrm{P}(X=3)\)

To find the probability of hitting the target exactly 3 times, we will use the Binomial Probability formula: \[\mathrm{P}(X=3) = \binom{8}{3}\left(\frac{2}{3}\right)^3\left(\frac{1}{3}\right)^{8-3}\] Now, let's compute the probability: \[\mathrm{P}(X=3) = \frac{8!}{3!(8-3)!}\left(\frac{2}{3}\right)^3\left(\frac{1}{3}\right)^{5}\] \[\mathrm{P}(X=3) \approx 0.129\] The probability of hitting the target exactly 3 times is about \(0.129\).

(b) \(\mathrm{P}(1< X \leq 6)\)

To find the probability of hitting the target between 1 and 6 times (not including 1), we will calculate the probabilities for each case (2, 3, ..., 6 hits) and sum them up. \[\mathrm{P}(1< X \leq 6)\] Using the Binomial Probability formula, we get: \[\mathrm{P}(1< X \leq 6) \approx \sum_{k=2}^{6} \binom{8}{k}\left(\frac{2}{3}\right)^k\left(\frac{1}{3}\right)^{8-k}\] \[\mathrm{P}(1< X \leq 6) \approx 0.346\] The probability of hitting the target between 1 and 6 times is about \(0.346\).

(c) \(\mathrm{P}(X>3)\)

To find the probability of hitting the target more than 3 times, we would usually have to calculate the probabilities for each case (4, 5, ..., 8 hits) and sum them up. However, we can save some work by computing \(\mathrm{P}(X \le 3)\) first and then using the complement rule: \(\mathrm{P}(X > 3) = 1-\mathrm{P}(X \le 3)\). The range for \(X \le 3\) includes cases for 0, 1, 2, and 3 hits. Using the Binomial Probability formula, we get: \[\mathrm{P}(X > 3) \approx 1 - \sum_{k=0}^{3} \binom{8}{k}\left(\frac{2}{3}\right)^k\left(\frac{1}{3}\right)^{8-k}\] \[\mathrm{P}(X > 3) \approx 0.851\] The probability of hitting the target more than 3 times is about \(0.851\).

Key concepts

Binomial Distribution

The binomial distribution is a type of probability distribution that outlines the likelihood of a given number of successes in a series of independent and identically distributed Bernoulli trials. In simpler terms, if you have a fixed number of repetitions of an experiment (like shooting at a target), and each experiment has two possible outcomes (success, i.e., hitting the target, or failure, missing the target), then you can use the binomial distribution to calculate the probability of achieving a certain number of successes. For example, in our exercise, with 8 shots and a 2/3 chance of hitting the target each time, we consider this a binomial setting with parameters \(n=8\) and \(p=\frac{2}{3}\).
This distribution is characterized by its use of the binomial coefficient, \(\binom{n}{k}\), and it forms the basis for the binomial probability formula.

Probability Theory

Probability theory is a branch of mathematics that deals with the likelihood of different outcomes. It's the foundation for understanding events that are governed by chance. When you're working with probabilities, like in our exercise, you use it to quantify the likelihood of an event occurring, such as hitting the target 3 times out of 8 shots.
In probability theory, the sum of probabilities of all possible outcomes equals 1. It helps us with tools to calculate the odds of complex scenarios by breaking down problems into simpler, manageable parts.

• The probability of an event can be expressed as a decimal or fraction between 0 and 1.
• A probability of 0 means the event is impossible, and 1 means it is certain.

For instance, the exercise uses probability calculations to determine the likelihood of hitting the target different numbers of times.

Random Variable

A random variable is a numerical value associated with the outcome of a random event. In statistical experiments like our target shooting exercise, a random variable is used to describe outcomes with a numeric value.
In the context of the binomial distribution, the random variable \(X\) represents the number of successes (hits) out of a certain number of trials. For instance, \(X=3\) means hitting the target exactly 3 times out of 8 shots.

• Discrete random variables can take on a countable number of distinct values.
• Our experiment is an example of a discrete random variable with a finite number of outcomes (0 to 8 hits).

This concept helps transform abstract probability problems into calculable outcomes.

Combinatorics

Combinatorics is a field of mathematics concerned with counting, arrangement, and combination of objects. It's essential when we need to calculate probabilities in scenarios where arrangements matter.
In the context of our exercise, we use combinatorics to determine the number of possible ways to achieve a specific number of successful shots — for instance, hitting the target exactly 3 times out of 8 attempts.
This is where the binomial coefficient, \(\binom{n}{k}\), comes in: it calculates the number of ways to choose \(k\) successes from \(n\) trials. It's crucial for calculating binomial probabilities, since different combinations can lead to the same number of successes.

• For example, \(\binom{8}{3}\) calculates how many ways we can achieve exactly 3 hits in 8 shots.
• These techniques simplify the process of calculating probabilities in numerous real-world situations, from card games to genetic probabilities.

Understanding combinatorics is key to mastering the calculation part of probability problems.