What is the probability that if five hats are distributed among five boxes that box $B_1$ has hat $H_1$ or hat $H_2$ but not both? There are $5$ different boxes $B_1, B_2, B_3, B_4, B_5$ and $5$ different hats $H_1, H_2, H_3 H_4, H_5$. The hats are to be distributed among the different boxes. Each box can accommodate all the hats.

A) What is the probability that $B_1$ has either $H_1$ or $H_2$ but not both?

B) If $B_1$ and $B_2$ have the hats $H_1$ and $H_2$ among themselves, in how many ways can you arrange the hats among the $5$ boxes?

Answer of A part is $2 \cdot 5^4-5^3/5^5=9/25$

Answer of B part is $4 \cdot 5^3$

Can anyone please explain this to me?
Each of the different ways of distributing the hats can be described as a 5-tuple, $(x_1,x_2,x_3,x_4,x_5)$, where $x_i$ represents which box hat $i$ was placed in.

For example, $(1,1,2,1,4)$ corresponds to box1 containing hats 1 2 and 4, box2 containing hat3, and box4 containing hat5.

From rule of product we know that there are $5^5$ possible such 5-tuples and therefore $5^5$ different ways to distribute the hats.

The problem statement does not explicitly state how we randomly distribute the hats, but a common assumption for problems like these is that each hat is distributed uniformly and independently at random amongst the boxes, implying that each of those $5^5$ possible ways of distributing the hats are equally likely to occur.  This is an important observation because it allows us to use counting techniques to continue the problem.

For the first question, finding the probability that the first box has hat1 or hat2 but not both (while also possibly having additional hats beyond those), we count how many 5-tuples correspond to such an arrangement.  Again, approach via rule of product breaking it down into steps as follows:


Pick whether it is hat1 or hat2 that is in box1  (2 options)
Whichever wasn't picked in the first step, pick which box it does go to remembering that it cannot be box1 (4 options)
Pick which box hat3 is in (5 options)
Pick which box hat4 is in (5 options)
Pick which box hat5 is in (5 options)


Applying the rule of product, we have then $2\cdot 4\cdot 5\cdot 5\cdot 5 = 8\cdot 5^3$ such arrangements of hats satisfying the phrase "exactly one of (but not both of) the first two hats are in box1."

Using what we know about finite equiprobable sample spaces then, the probability of this happening is the ratio of the number of favorable outcomes divided by the total number of outcomes.  I.e. a probability of $\frac{2\cdot 4\cdot 5^3}{5^5}=\frac{8}{25}$.

This contradicts your answer in the original post which might have caused some of your confusion.  An answer of $\frac{9}{25}$ actually corresponds to the probability of at least one of the first two hats being in the first box, including the possibility of both.  To have calculated this, we could approach similarly to above, having first broken into cases based on whether exactly one or both of the first two hats were in the first box, giving a total number of valid sequences being $2\cdot 4\cdot 5^3+1\cdot 1\cdot 5^3$ and the probability then as $\frac{2\cdot 4\cdot 5^3+1\cdot 1\cdot 5^3}{5^5}=\frac{2\cdot 4+1}{5^2}=\frac{9}{25}$.  Alternatively, you could have calculated this using inclusion exclusion arriving at the same answer.

The second part of the problem is very similar to the first.  The steps for rule of product being:


Pick location of hat 1
Pick location of hat 2
Pick location of hat 3
Pick location of hat 4
Pick location of hat 5



   With 2 options for each of the first two steps and 5 options for each of the remaining steps, that gives the total number of valid arrangements as $2\cdot 2\cdot 5\cdot 5\cdot 5=4\cdot 5^3$ valid sequences.  Taking the ratio then, this gives the probability as $\frac{4\cdot 5^3}{5^5}=\frac{4}{25}$
My reasoning seems too simplistic and is probably flawed but does give the solution to the first problem. The chance of hat1 not being in box1 is $\frac 4 5$ and likewise for hat2. $\frac 4 5 \times \frac 4 5 = \frac {16} {25}$. Thus the chance of one or both being in box1 is $\frac {25} {25} - \frac {16} {25} = \frac 9 {25}$. The chance of both being in box1 is $ \frac 1 5 \times \frac 1 5 = \frac 1 {25}$. So for only one of them we have $\frac 9 {25} - \frac 1 {25} = \frac 8 {25}$.