the prime digits are=》2,3,5,7
As 2 is fixed at first place (2n-1)places are filled using this 4 digit(2,3,5,7)
2_ _ _ _ ........_ _ _ _
as sum of two cons digit is prime hence
2+3=5, 2+5=7, 2 +7=9(not prime), 3+5=8(not prime), 3+7=10 (not prime), 5+7=12( not prime)
hence only (2,3) and(2,5) is required cons digit
now, total way to fill(2n-1)using 4 different digit is 4^(2n-1){sample space}
Fix 2 at alternative position because both pair starts from 2 [(2,3) (2,5)]
2_2_2_2_2_2...._2_2_
now gap place can filled using either 3 or 5[ because (2+3=5) or (2+5=7) prime no.]
Hence ,total required way to fill is 2^n
Required probability = 2^n/4^(2n-1)=4/2^3n
