First slide

Introduction to probability

Question

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

Moderate

A

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

$4 \times 2^{- 2 n}$
C

$2^{- 3 n}$
D

none of these

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 favorable number of cases is 2ⁿ. Therefore, probability is
$\frac{2^{n}}{4^{2 n - 1}} = 2^{n} 2^{- 4 n + 2} = 2^{2} 2^{- 3 n} = \frac{4}{2^{3 n}}$

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