# A 2n digit number starts with 2 and all its digits are prime, then the probability that the sum of all 2 consecutive digits of the number is prime is

Question : A 2n digit number starts with 2 and all its digits are prime, then the probability that the sum of all 2 consecutive digits of the number is prime is

(A) $4\times 2^{3n}$

(B) $4\times 2^{-3n}$

(C) $2^{3n}$

(D) none of these

Answer : option (B)

Solution :

The prime digits are 2, 3, 5, 7. If we fix 2 at first place, then other 2n-1 places are filled by all four digits. So the total number of cases is $4^{2n-1}$

Now, sum of 2 consecutive digits is prime when consecutive digits are (2,3) or (2, 5) Then 2 will be fixed at all alternative places

So favourable number of cases is $2^{n}$.

Therefore, probability is

$\dfrac{2^{n}}{4^{2n-1}}=2^{n}2^{-4n+2}=2^{2}2^{-3n}=\dfrac{4}{2^{3n}}$
Tags: No tags