Problem 4 - Arbeitsbuch Mathematik für Ingenieure

Question

Ein Glücksrad hat die Felder \(1,2,3\) und 4, die alle mit der gleichen Wahrscheinlichkeit (also \(0.25\) ) auftreten. Es wird solange gedreht, bis zum ersten Mal eine 4 erscheint, höchstens jedoch fünfmal. Die Zufallsvariable \(X\) beschreibe die Anzahl der Versuche. a) Geben Sie die Verteilung von \(X\) in einer Tabelle an. b) Skizzieren Sie die Verteilungsfunktion. c) Wie groß ist die Wahrscheinlichkeit, mindestens zweimal, jedoch höchstens viermal zu drehen?

Step-by-Step Solution

Answer

P(2 ≤ X ≤ 4) = 0.421875

Step 1: Understand the problem

A Glücksrad (spinning wheel) has four fields numbered 1, 2, 3, and 4, each appearing with the same probability (0.25). The wheel is spun until the number 4 appears for the first time, but no more than five times. The random variable X represents the number of spins. Determine the distribution, sketch the cumulative distribution function, and find the probability of spinning the wheel between 2 and 4 times inclusive.

Step 2: Determine probabilities for each spin outcome of X

Since each field has a probability of 0.25 of appearing, the probability of getting a 4 on the first try is 0.25. For subsequent tries, the probability of rolling non-4s until the 4 appears can be computed using the geometric distribution:

Step 3: Calculate specific probabilities

The probability mass function (pmf) for a geometric distribution with a success probability of p = 0.25 is:

Step 4: Set up and fill in the probability distribution table

a) Create a table with X values from 1 to 5 and calculate the probabilities:

Step 5: Draw the cumulative distribution function (CDF)

b) The CDF can be sketched by plotting cumulative probabilities, summing the probabilities from the start up to each X value. Draw a plot of P(X ≤ k) for k = 1 to 5.

Step 6: Compute the requested probability

c) Calculate the probability of spinning at least twice but no more than four times by summing the relevant probabilities: P(2 ≤ X ≤ 4) = P(X=2) + P(X=3) + P(X=4).

Key Concepts

Probability Distribution

A probability distribution shows how probabilities are assigned to different outcomes in a random experiment. In this case, the number of spins until a 4 appears for the first time forms a geometric distribution. Each spin has a success probability of 0.25 (since there are four fields, each equally likely). The probability distribution tells us how likely each possible outcome (from 1 to 5 spins) is.
For example, the probability of getting a 4 on the first spin is 0.25. The table below outlines the probability for each number of spins up to five:
- P(X = 1) = 0.25
- P(X = 2) = 0.1875
- P(X = 3) = 0.140625
- P(X = 4) = 0.10546875
- P(X = 5) = 0.421875
This way, you can see the distribution of probabilities across different outcomes.

Cumulative Distribution Function

The cumulative distribution function (CDF) gives the probability that a random variable is less than or equal to a certain value. In simpler terms, it sums up the probabilities of all outcomes up to and including a specific point.
For the given exercise, the CDF for the number of spins can be plotted as follows:
- P(X ≤ 1) = 0.25
- P(X ≤ 2) = P(X = 1) + P(X = 2) = 0.25 + 0.1875 = 0.4375
- P(X ≤ 3) = P(X ≤ 2) + P(X = 3) = 0.4375 + 0.140625 = 0.578125
- P(X ≤ 4) = P(X ≤ 3) + P(X = 4) = 0.578125 + 0.10546875 = 0.68359375
- P(X ≤ 5) = 1 (since the sum of all probabilities in a distribution is 1)
This cumulative values plot helps understand how the probabilities build up until we reach a certain number of spins.

Random Variable

A random variable is a quantity whose value is determined by the outcome of a random phenomenon. In our problem, the random variable X represents the number of spins taken until a 4 appears.
Random variables can be discrete or continuous. Here, X is discrete since it can only take integer values (1, 2, 3, 4, and 5). Understanding the behavior and the expected value of the random variable is crucial to analyzing the probability distribution.

Geometric Probability

Geometric probability deals with the number of trials needed to get the first success in repeated independent Bernoulli trials. Each trial has two possible outcomes (success or failure), and the trials are independent.
In the problem, each spin of the wheel is a trial where spinning a 4 is a success with a probability of 0.25, and not spinning a 4 is a failure with a probability of 0.75.
The general formula for the probability of the first success occurring on the k-th trial in a geometric distribution is given by:
\[ P(X = k) = (1 - p)^{(k-1)} \times p \]
where p is the success probability (0.25 in our case). This captures the essence of geometric probability and how to calculate it.

Probability Mass Function (pmf)

The probability mass function (pmf) provides the probability that a discrete random variable is exactly equal to some value. For a geometric distribution, the pmf is:
\[ P(X = k) = (1 - p)^{k-1} \times p \]
For our specific problem where the probability p of spinning a 4 is 0.25, the pmf can be used to find how likely it is to get a 4 on each spin from 1 to 5.
Using this function:
\[ P(X = 1) = 0.25 \]
\[ P(X = 2) = 0.75 \times 0.25 = 0.1875 \]
\[ P(X = 3) = 0.75^2 \times 0.25 = 0.140625 \]
\[ P(X = 4) = 0.75^3 \times 0.25 = 0.10546875 \]
\[ P(X = 5) = 0.75^4 \times 0.25 = 0.421875 \]
This function helps us easily calculate the probabilities for each specific number of spins.