InterviewSolution
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InterviewSolution
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A `2n`digit number starts with 2 and all its digits are prime, then the probabilitythat the sum of all 2 consecutive digits of the number is prime is`4xx2^(3n)`b. `4xx2^(-3n)`c. `2^(3n)`d. none of theseA. `4 xx 2^(-3n)`B. `4 xx 2^(-3n)`C. `2^(-3n)`D. none of these |
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Answer» Correct Answer - B 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,...,...,2,):}` So favorable number of cases is `2^(n)`. Therefore, probability is `(2^(n))/(4^(2n-1)) = 2^(n) 2^(-4n+2) = 2^(2) 2^(-3n) = (4)/(2^(3n))` |
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