$\dfrac{30}{60} \cdot \dfrac{20}{30}+\dfrac{20}{60} \cdot \dfrac{30}{40} =\dfrac{7}{12}$

We will reach this formula by considering the complete sequence of 60 candies.

Any scenario that satisfies the given question will match one of the following two:

1. The $60^{\text{th}}$ candy in this sequence is green, and some blue candy exists before this and after the last red candy.

2. The $60^{\text{th}}$ candy is blue, and some green candy exists before this and after the last red candy.

In the first scenario, let $G$ be the event that the $60^{\text{th}}$ candy is green. Let $B^*$ be the event that there is a blue candy before that but after the last red. Since these two events are not independent, we want to calculate:

$P(G \cap B^*) = P(G) \cdot P(B^* | G)$

$P(G) = 30/60$ because the $60^{\text{th}}$ candy will be one of 30 greens, and there are a total of 60 choices.

$P(B^* | G)$ is the probability of finding a blue candy after the last red, given that the $60^{\text{th}}$ candy is green. Now remove all green candies from consideration, there are 30 candies (10 red & 20 blue). Imagine a random sequence of these 30 candies (it might be scattered anywhere from position 1 to $59$, but still, it is a random sequence of 30 candies only). The probability that the $30^{\text{th}}$ one of these is blue is $20/30$.

Thus $P(G \cap B^*) = (30/60) \cdot (20/30)$.

Similarly, for the second scenario, let $B$ be the event that the $60^{\text{th}}$ candy is blue and let $G^*$ be the event that there is a green candy before that but after the last red. We want to calculate $P(B \cap G^*) = P(B) \cdot P(G^* | B) = (20/60) \cdot (30/40)$

Since these two scenarios are mutually exclusive, the probability of union is the sum of these probabilities.

Hence the answer: $1/3 + 1/4 = 7/12$

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Not convinced? Here is a simulation script that shows the probability converging to 7/12.

Colab Script