(1) Exactly two pairs are taken as C5 (1) C4 (2) C2 (1) C2 (1) =120.

∴ of the four shoes, two are just double P = C5 (1) C4 (2) C2 (1) C2 (1)/c10 (4) =12/.

(2) The event "at least two of the four shoes are in pairs" includes the events "exactly two shoes are in pairs" and "exactly four shoes are in pairs", and the method to get exactly two pairs is C5 (1) C4 (2) C2 (1) C2 (1) = 650.

At least two of the four shoes are in pairs, p = C5 (1) C4 (2) C2 (1) C2 (1)+C5 (2)/c10 (4) =/kloc-0.

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