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Sophie is a dog who loves to play catch. Unfortunately, she isn't very good at this, and the probability that she catches a ball is only \(0.1 .\) Let \(x\) be the number of tosses required until Sophie catches a ball. a. Does \(x\) have a binomial or a geometric distribution? b. What is the probability that it will take exactly two tosses for Sophie to catch a ball? c. What is the probability that more than three tosses will be required?

Short Answer

Expert verified
a. x has a geometric distribution. b. The probability that it will take exactly two tosses for Sophie to catch a ball is 0.09 (or 9%). c. The probability that more than three tosses will be required for Sophie to catch a ball is 0.729 (or 72.9%).

Step by step solution

01

Determine the Distribution

For x to have a binomial distribution, we would need a fixed number of trials (n) and the probability of success (p) in each trial. However, in this case, we are looking for the number of trials (tosses) it takes for Sophie to catch a ball, which is not fixed. Hence, x has a geometric distribution. A geometric distribution is used to model the number of trials required for the first success. In this case, success represents Sophie catching a ball. The probability of success in each trial is p = 0.1. So, x follows a geometric distribution with parameter p = 0.1. a. x has a geometric distribution.
02

Calculate the Probability of Catching a Ball in Exactly Two Tosses

For a geometric distribution, the probability mass function (PMF) for exactly n trials until the first success is given by: \(P(X=n) = (1-p)^{(n-1)}p\) In this case, n = 2, and p = 0.1: \(P(X=2) = (1-0.1)^{(2-1)}(0.1) = 0.9 \times 0.1 = 0.09\) b. The probability that it will take exactly two tosses for Sophie to catch a ball is 0.09 (or 9%).
03

Calculate the Probability of More Than Three Tosses Required

Since x has a geometric distribution, we need to calculate the probability that more than three tosses are required, i.e., P(X > 3). Using the complement rule for probabilities, we can compute this value as follows: \(P(X > 3) = 1 - P(X \leq 3)\) Now, we need to calculate the probability that either of the first three tosses results in a caught ball (P(X = 1) ∪ P(X = 2) ∪ P(X = 3)). Since the geometric distribution is memoryless, these events are disjoint (non-overlapping), so we can add their probabilities: \(P(X\leq 3) = P(X=1) + P(X=2) + P(X=3)\) Using the geometric PMF: \(P(X=1) = 0.1\) \(P(X=2) = 0.09\) (from part b) \(P(X=3)=(1-0.1)^{(3-1)}(0.1) = 0.9^2 \times 0.1 = 0.081\) Combining all probabilities: \(P(X\leq 3) = 0.1 + 0.09 + 0.081 = 0.271\) Now, we can compute the probability of more than three tosses required using the complement rule: \(P(X > 3) = 1 - 0.271 = 0.729\) c. The probability that more than three tosses will be required for Sophie to catch a ball is 0.729 (or 72.9%).

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Key Concepts

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

Probability Mass Function
Understanding the probability mass function (PMF) is crucial when dealing with discrete random variables like the number of tosses before Sophie catches a ball. In our scenario, the PMF, denoted as \( P(X=n) \), describes the probability that the random variable \( X \) takes on the value \( n \). This is the number of trials it takes until the first success occurs. It's important to know that the PMF applies only to discrete variables; for continuous random variables, we use the probability density function (PDF).

For a geometric distribution, the PMF is given by \( P(X=n) = (1-p)^{(n-1)}p \), where \( p \) is the probability of success on each trial. When analyzing our example with Sophie, if we want to find out the probability that she catches the ball on the second toss (\( n=2 \)), we plug this value into our PMF equation to get \( P(X=2) \) which, as calculated, yields 0.09, or 9%.
Complement Rule
The complement rule is a fundamental concept in probability theory that helps us calculate the likelihood of an event not occurring by subtracting the probability of the event occurring from 1. Simply put, if the probability of Sophie catching a ball on any toss is \( P \), then the probability of her not catching it on that toss is \( 1 - P \).

More formally, for any event \( A \), \( P(A') = 1 - P(A) \), where \( A' \) denotes the complement of \( A \). We applied this rule to determine the probability of Sophie not catching the ball in three tosses or less (\( P(X > 3) \)). By subtracting the probability of the opposite event (catching the ball within three tosses) from 1, we obtained the result of 0.729 or 72.9%. This is useful in scenarios where computing the probability of the complementary event is easier than directly calculating the probability of the target event.
Binomial Distribution
It's important to differentiate a geometric distribution from a binomial distribution, as both relate to the probability of a series of independent and identically distributed Bernoulli trials, which are experiments with only two possible outcomes: success (e.g., Sophie catches the ball) or failure. The binomial distribution, however, measures the probability of achieving a specific number of successes (\( k \)) in a fixed number (\( n \)) of trials, with the probability of success on each trial being \( p \).

The probability of achieving exactly \( k \) successes in a binomial setting is computed using the formula \( P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} \), where \( \binom{n}{k} \) denotes the binomial coefficient. In contrast to the geometric distribution, which focuses on the number of trials until the first success, the binomial distribution considers the number of successes within a predetermined number of trials. Understanding the distinctions between these two distributions is key for correctly modeling various types of probability questions.

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Most popular questions from this chapter

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