The average of 3 prime numbers tying between 4th to7th is $\dfrac{{191}}{8}.$ The greatest possible difference between any two out of the 3 prime numbers is.

Hint : We know that the average of a fore integer is the sum of integers divided by no of integers. Here, The average of any there prime number between 4th and 7th is $\dfrac{{191}}{3}$ which is a fraction

Complete step-by-step answer:
In $\dfrac{{191}}{3}$, the 3 is the number of prime number taken
So, we can say that the sum of a prime number is 191 which is a numeration.
Let’s say the prime number between 47 and 74 is 53,59,61,67,71,73.
So, for getting 191 as a sum. There are only two ways
1). $53 + 67 + 71 = 191$
2). $59 + 61 + 71 = 191$
So, arranging these no we get,
53,59, 61,67 ,71
So, greatest possible difference will be $71 - 53 = 18 = {\text{Answer}}{\text{.}}$

So, our answer is 18.

Formula used:
${\text{Average}}\;{\text{ = }}\dfrac{{{\text{sum}}\;{\text{of}}\;{\text{integer}}}}{{{\text{number}}\;{\text{of}}\;{\text{integer}}}}.$

Note : In this type of question always try to look at the range of numbers given and use hit and trial methods to solve these types of problems.