Initially, I also had the same impression, but when you look at it carefully,
Number of choices for 1st multicast group address: $2^{28}$
Number of choices for 2nd multicast group address: 1(we choose only the address chosen by group 1)
Calculating the probability: ($2^{28}$ / $2^{28}$) * (1 / $2^{28}$)
Which gives us: 1/($2^{28}$)
Still confused?
According to you the answer should be: (1/$2^{28}$)*(1/$2^{28}$)
This unfortunately is the probability of both the multicast groups selecting the same(only a particular, say, 224.1.1.1) group address.
But, we have $2^{28}$ choices: Therefore multiplying your answer with $2^{28}$ gives us the correct result.