Question

\(60 \%\) der Studenten, die das Physikstudium mit dem Ziel „DiplomPhysiker " beginnen, schließen ihr Studium erfolgreich ab. Wie groB ist die Wahrscheinlichkeit, daB in einer Gruppe von zehn zufällig ausgewählten Physikstudenten des 1. Semesters acht das Physikdiplom erhalten?

Short answer

0.121

Step-by-step solution

Understanding the Problem

We need to find the probability that exactly 8 out of 10 students will successfully graduate, given that the success rate is 60%. This is a binomial probability problem.

Identify Parameters

Identify the parameters for the binomial distribution:- Number of trials () is 10 (since there are 10 students)- Probability of success () is 0.60 (since 60% of students graduate)- Number of successes () is 8 (we want to know the probability that exactly 8 students graduate)

Binomial Probability Formula

Use the binomial probability formula: \[P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}\]Where:- \( \binom{n}{k} \) is the number of combinations of n items taken k at a time- \( p \) is the probability of success- \( 1 - p \) is the probability of failure

Calculate Combinations

Calculate the combinations term, \( \binom{10}{8} \):\[\binom{10}{8} = \frac{10!}{8!(10-8)!} = \frac{10!}{8!2!} = 45\]

Calculate Probability

Substitute the values into the binomial formula:\[P(X = 8) = 45 \cdot (0.60)^8 \cdot (0.40)^2\]Calculate each term separately:- \( (0.60)^8 = 0.0168 \)- \( (0.40)^2 = 0.16 \)Combine them:\[P(X = 8) = 45 \cdot 0.0168 \cdot 0.16 = 0.12096\]

Key concepts

Binomial Distribution

The concept of *binomial distribution* is crucial here. It's used to determine the probability of a given number of successes in a fixed number of trials. Here, each trial is an attempt to graduate, and there are 10 such trials (students). The binomial distribution assumes that each trial is independent, meaning the outcome of one student doesn't affect the others. If we denote the number of trials by \(n\), the number of successful outcomes we are interested in by \(k\), and the probability of success by \(p\), the distribution helps calculate the likelihood of having exactly \(k\) successes out of \(n\) trials.
This concept breaks down complex probability into simpler steps, making it easy to compute probabilities for multiple events.

Probability of Success

In binomial problems, we need to know the *probability of success* for a single trial. Here, the success rate means graduating, which is 60%, or 0.60.
We denote this probability by \(p\). For the problem, \(p = 0.60\). The complement, or the probability of failure, is \(1 - p\), which equals 0.40 in this case. This small yet important detail ensures we account for the chance of both successes and failures while computing the overall probability.

Combinations

The term *combinations* refers to the number of ways to choose a subset of items from a larger set, without considering the order. In our scenario, it means choosing 8 successful students out of 10. This value is represented mathematically by \(\binom{n}{k}\), where \(n\) is the total number of students and \(k\) is the number of successful graduates.
We calculate it using the formula:
\[\binom{10}{8} = \frac{10!}{8!(10-8)!} = \frac{10!}{8!2!} = 45\]
This means there are 45 different ways to arrange 8 graduating students among 10.

Probability Formula

Finally, we use the *binomial probability formula* to put everything together:
\[P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}\]
This formula helps us find the probability of getting exactly \(k\) successes in \(n\) trials, given the probability of success \(p\). For our exercise, we substitute the known values:
\[P(X = 8) = 45 \cdot (0.60)^8 \cdot (0.40)^2\]
Next, calculate each part separately:
\( (0.60)^8 = 0.0168 \)
\( (0.40)^2 = 0.16 \)
Finally, multiply them together:
\[P(X = 8) = 45 \cdot 0.0168 \cdot 0.16 = 0.12096\]
Thus, there's an approximately 12.1% chance that exactly 8 out of 10 students will successfully graduate. This step-by-step breakdown clarifies how various pieces fit into the overall probability.