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$\text{This is the same as the probability you've eaten 2 chocolates in 6 candies, and eat the 3rd chocolate on the 7th}\\ P[\text{eaten 2 chocolates in 6 candies}] = \dfrac{\dbinom{3}{2}\dbinom{5}{4}}{\dbinom{8}{6}} = \dfrac{15}{28}\\ P[\text{7th candy eaten is chocolate |2 chocolates already eaten}] = \dfrac 1 2\\ P = \dfrac{15}{28}\dfrac 1 2 = \dfrac{15}{56}$