E = तीसरी उछाल पर संख्या 4 प्रकट होना तथाF पहली दो उछालों पर क्रमश: 6 तथा 5 प्रकार होना
= (1,1, 4), (1, 2, 4), (1, 3, 4), … (1, 6, 4)
= (2, 1, 4), (2, 2, 4), (2, 3, 4), … (2, 6, 4)
= (3, 1, 4), (3, 2, 4), (3, 3, 4), … (3, 6, 4)
= (4,1, 4), (4, 2, 4), (4, 3, 4), … (4,6, 4)
= (5, 1, 4), (5, 2, 4), (5, 3, 4), … (5, 6, 4)
= (6,1, 4), (6, 2, 4), 6, 3, 4),… (6, 6, 4)
= 36 परिणाम
तथा F = {6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6)} = 6 परिणाम
∴ E ∩ F = {6, 5, 4}
