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­Here is my approach

First 3 rolls have 8 possibilities
(1)__(2)__(3)
_E___E___E_
_E___E___O_ (a)
_E___O___E_ (b)
_E___O___O_
_O___E___O_
_O___E___E_ (c)
_O___O___O_
_O___O___E_

2 of the first 3 rolls are even => Remove the red and only the green ones remain

2 of the last 3 rolls are even. Then we have
(1)__(2)__(3)__(4)__(5)
_E___E___O___E____E (a)
_E___O___E___O____E (b)
_E___O___E___E____O (b)

_O___E___E___O____E (c)
_O___E___E___E____O (c)


So the 3rd roll is even in 4 out of 5 possibilities
=> Probability: \(\frac{4}{5}\)­
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