a)

The number of outcomes in $S$ is given by $6^4=1296$

b)

There are two ways to place vowels in both the first slot and last slot, and six ways to place any of the letters in each of the two center slots. Hence, there are $2^2×6^2=144$ outcomes in $𝒜$, so $P(𝒜)=\frac{144}{1296}=0.111$

c)

There are $6^{(4)}=\frac{6!}{2!}=360$ outcomes in $ℬ$, so $P(ℬ)=\frac{360}{1296}=0.278$

d)

For $𝒜∩ℬ$, the word must either start with E and end with I or start with I and end with E. Then, the remaining 4 characters can be distributed uniquely amongst the remaining two slots in $4^{(2)}=12$ ways. Thus, there are $2×12=24$ outcomes in $𝒜∩ℬ$, so $P(𝒜∩ℬ)=\frac{24}{1296}=0.0185$

Finally, by the sum rule, $P(𝒜∪ℬ)=P(𝒜) + P(ℬ) - P(𝒜∩ℬ) = \frac{144}{1296} + \frac{360}{1296} - \frac{24}{1296} = \frac{480}{1296}=0.370$

e)

No; $P(𝒜)P(ℬ)=\frac{144}{1296}×\frac{360}{1296}=\frac{5}{162}≠P(𝒜∩ℬ)$, and so they are not independent by definition.