**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 **:

2 |

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

2 | 2 | 2 | 2 |

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}}