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
`4xx2^(3n)`
b. `4xx2^(-3n)`
c. `2^(3n)`
d. none of these
A. `4 xx 2^(-3n)`
B. `4 xx 2^(-3n)`
C. `2^(-3n)`
D. none of these
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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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