<p>A \(2n\) digit number starts with 2 and all its digits are prime, then the probability that the sum of any two consecutive digits of the number is prime is</p>
Step-by-Step Solution
Key Concept: A number satisfies the condition if every pair of consecutive digits sums to a prime. Since the first digit is 2, we must find which prime digits (2,3,5,7) can follow each prime digit such that their sum is prime, then count valid sequences.
<p><strong>Step 1:</strong> Identify valid transitions. Prime digits are {2,3,5,7}. For consecutive digits a,b: a+b must be prime.</p><p><strong>Step 2:</strong> Check which digits can follow each prime:</p><ul><li>After 2: 2+2=4(✗), 2+3=5(✓), 2+5=7(✓), 2+7=9(✗) → can use {3,5}</li><li>After 3: 3+2=5(✓), 3+3=6(✗), 3+5=8(✗), 3+7=10(✗) → can use {2}</li><li>After 5: 5+2=7(✓), 5+3=8(✗), 5+5=10(✗), 5+7=12(✗) → can use {2}</li><li>After 7: 7+2=9(✗), 7+3=10(✗), 7+5=12(✗), 7+7=14(✗) → can use {}</li></ul><p><strong>Step 3:</strong> From any odd prime (3,5,7), only 2 can follow. From 2, we have 2 choices (3 or 5). Once we reach 3 or 5, we must return to 2. From 7, we cannot proceed.</p><p><strong>Step 4:</strong> Valid sequences starting with 2: After 2, choose 3 or 5 (2 ways). Then forced to 2. This cycles. For a 2n-digit number, valid sequences are of form: 2→{3 or 5}→2→{3 or 5}→...→2, giving 2^(n-1) valid sequences.</p><p><strong>Step 5:</strong> Total possible 2n-digit numbers with all prime digits = 4^(2n-1) (first digit is 2, remaining 2n-1 digits chosen from {2,3,5,7}).</p><p><strong>Step 6:</strong> Probability = 2^(n-1)/4^(2n-1) = 2^(n-1)/2^(4n-2) = 2^(n-1-4n+2) = 2^(1-3n) = 2/(8^n)</p><p>∴ Answer: B</p>
Correct Answer: B