Problem 110

To find the total number of ways to distribute 5 distinct books to 4 students, where each book can be given to any of the students, we use the multiplication rule of counting. There are 4 choices for the recipient of each book, leading to a total of $4^5$ different combinations.

Next, we determine the number of distributions where each student receives at least one book. To ensure each student gets at least one book, we must prevent any student from getting no books at all. We can use the inclusion-exclusion principle to avoid overcounting the distributions that leave one or more students without a book.

The idea is to calculate the total ways of distributions that leave out at least one student, and then subtract that from the total possible distributions:

1. First, we calculate the number of ways to exclude one particular student and distribute all five books among the remaining three students. There are 3 choices for each of the 5 books, which gives $3^5$ combinations.

2. However, there are 4 students, so to exclude any of the students, we have $4 \times 3^5$ combinations.

3. But this overcounts the distributions that exclude two students because such distributions are subtracted once for each student they exclude. Thus, we need to add back the number of ways to distribute the books among two students. There are $2^5$ combinations for two students and $\binom{4}{2}$ ways to choose which two students get the books. So, we add $\binom{4}{2} \times 2^5$.

4. Finally, if all 5 books go to one student, there are 4 ways for this to happen (one for each student), which we must also subtract given that it's been added one too many times in the previous step.

Putting it all together, we calculate the number of valid distributions where no student is left out:

$$\text{Valid distributions} = 4^5 - 4 \times 3^5 + \binom{4}{2} \times 2^5 - 4$$

Now, using the combination formula $\binom{n}{r} = \frac{n!}{r!(n-r)!}$ and factorial notation:

$$\text{Valid distributions} = 4^5 - 4 \times 3^5 + \frac{4!}{2!(4-2)!} \times 2^5 - 4$$

Plugging in the values, we get:

$$\text{Valid distributions} = 1024 - 4 \times 243 + 6 \times 32 - 4$$
$$\text{Valid distributions} = 1024 - 972 + 192 - 4$$
$$\text{Valid distributions} = 240$$

Now, the probability that each student receives at least one book is the number of valid distributions divided by the total number of combinations:

$$P = \frac{\text{Valid distributions}}{4^5} = \frac{240}{1024} = \frac{15}{64}$$

Therefore, the correct answer is:

$$P = \boxed{\frac{15}{64}}$$