Total two-digit numbers is 90
So, total number of cases =C190=90
Now, 2n−2=(3−1)n−2
=C0n3n−C1n⋅3n−1+…+(−1)n−1⋅Cn−1n3+(−1)n⋅Cnn−2
=3(3n−1−n3n−2+…+(−1)n−1⋅n)+(−1)n−2
Here, the term (−1)n−2 should be multiple of 3.
So, (2n−2) is multiple of 3 only when n is odd.
There are total 45 odd two-digit numbers.
Hence, the required probability =90C145=9045=21