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Total ways to chose 3 actions figures is 10C3 = 120
3 actions figures can be arranged in 3! ways = 6
For each of these combinations the thrid figure can be one amoung 8
so we have 6*8= 48 ways
probability that any two of the action figures are 1 and 2 = 48/120 = 2/5
probability that the set of three does not contain two of them = 1-2/5 = 3/5
Please post the official answer.
Explain how you gor 5/8. May then the problem will become clear.