Let, x1, x2, x3, x4 are the results of the four dice respectively.
Here, x1 + x2 + x3 + x4 = 22 → eqn 1
4 <= x1, x2, x3, x4 <= 6 → $a$
[ if anyone of the dice shows less than 4, the sum of the four can’t be 22, :) ]
Let,
x1 = 6 – y1
x2 = 6 – y2
x3 = 6 – y3
x4 = 6 – y4
Therefore,
y1 = 6 – x1
y2 = 6 – x2
y3 = 6 – x3
y4 = 6 – x4
[ Here, form a, 0 <= y1, y2, y3, y4 <= 2 ]
putting x1, x2, x3, x4 in eqn 1 ---
6 – y1 + 6 – y1 + 6 – y1 + 6 – y1 = 22
y1 + y2 + y3+ y4 = 2 → eqn 2
No. of solutions of eqn 2 = (n + r – 1 C r – 1)
= 2 + 4 – 1 C 4 – 1
= 5C2
= 10
Therefore, The probability that the sum of the results being 22 = 10 / ( 6 ^ 4)
= 10 / 1296
So, X = 10.