3. (20 points) At the Faculty of Journalism of the University of Enchanted Commonwealth, there are 5 targeted places: 2 for the daytime department and 3 for the evening department. Four chickens have applied for these places: three black and one white. A place is reserved for each chicken immediately after they submit their application, and the order in which the chickens submit their applications is strictly random. If there are free places in both departments, each chicken chooses either with equal probability. Find the probability that at least one black chicken will be admitted to the daytime department.
Answer: 0.922.

Solution. The faculty has a total of 5 places, to which 4 applicants are applying. Since all applicants will be admitted to the faculty, the outcome is determined by the selection of those who will be admitted to the daytime department. This can be either two applicants (who can be chosen in six ways) or one applicant (who can be chosen in four ways); thus, there are a total of 10 outcomes. Let's find the probabilities of these outcomes.

Let's fix the order in which the chickens submit their applications. Let this be $K_{1} K_{2} K_{3} K_{4}$. Thus, the chicken $K_{1}$ will make the choice first, followed by $K_{2}$, and so on. Since there are no restrictions on the preferences of the applicants, the choice between the daytime and evening departments is equally probable and equal to $\frac{1}{2}$, if technically possible (not all options are available for the chickens making the last choices - due to limitations on the number of places both in the daytime and evening departments).

![](https://cdn.mathpix.com/cropped/2024_05_06_de56aaa25ca847a38066g-2.jpg?height=554&width=1447&top_left_y=1856&top_left_x=333)

On the diagram, double lines leading from a node to the left denote the choice of the daytime department, and single lines leading from a node to the right denote the evening department. Solid lines represent technically feasible cases, dashed lines represent impossible variants. The specific chicken making the choice at a particular level is noted in the left column. If a certain variant requires a choice by $n$ chickens, the probability of such a variant is $\frac{1}{2^{n}}$. Note that, for example, for variant $i_{1}$ to be realized, only the first two chickens need to make a choice - the third and fourth have no choice but to choose the evening department. The probabilities of each possible variant are given at the bottom of the diagram under the names of the variants.

The chickens submit their applications in a random order, so when the solution mentions, for example, one chicken out of four, it can be the first, second, third, or fourth among those submitting applications; and secondly, if we are interested in a specific chicken rather than a random one, we need to understand how many ways we can choose the needed chicken from the available ones. The answer will be the product of the corresponding probabilities.

The event "at least one black chicken will be admitted to the daytime department" is the opposite of the event "no black chickens will be admitted to the daytime department." At least one chicken must be on the daytime department (since there are 3 places in the evening department and 4 chickens in total), and since we have only one white chicken, it will be the one admitted to the daytime department. The probability of choosing exactly one (white) chicken out of the four available is $\frac{1}{4}$.

Now let's find the probability that "exactly one chicken will be admitted to the daytime department." On the diagram, these are events $i_{4}, i_{7}, i_{9}$, and $i_{10}$ - when the daytime department is chosen by chicken $K_{1}, K_{2}, K_{3}$, and $K_{4}$, respectively. The probability of their realization is $\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{8}=\frac{5}{16}$.

From this, we get the answer: $1-\frac{5}{16} \cdot \frac{1}{4}=\frac{59}{64} \approx 0.922$.