Suppose we are given $n$ keys, $m$ hash table slots, and two simple uniform hash functions $h_1$ and $h_2$. Further suppose our hashing scheme uses $h_1$ for the odd keys and $h_2$ for the even keys. What is the expected number of keys in a slot?

Let $n_{\text{odd}}$ be the number of odd keys and $n_{\text{even}}$ be the number of even keys, such that:

Under the Simple Uniform Hashing Assumption (SUHA):

• Any key hashed using $h_1$ is equally likely to hash to any of the $m$ slots with probability $\frac{1}{m}$.

• Any key hashed using $h_2$ is equally likely to hash to any of the $m$ slots with probability $\frac{1}{m}$.

For a specific slot $j$ ($1 \le j \le m$), let the indicator random variable $I_i = 1$ if key $i$ hashes to slot $j$, and $0$ otherwise. Then $E[I_i] = \frac{1}{m}$ for every key $i$, whether it is odd or even.

The total number of keys in slot $j$ is $X_j = \sum_{i=1}^n I_i$.

By the linearity of expectation:

Hence, the expected number of keys in any slot is $\frac{n}{m}$.

Therefore, option B is correct.