Q1：

　　(i)sixdefenceplayersarechosenfromtwentyfivehockeyplayers,thepossibleisC(25,6),then,fiveforwardplayersarechosenfromtheothernineteen,possibleisC(19,5)；

　　so,therecanbeC(25,6)*C(19,5)teams;

　　(ii)sixdefenceplayersarechosenfromthetenavailable,thepossibleisC(10,6),and,fiveforwardplayersarechosenfromthefifteenavailable,possibleisC(15,5)；

　　so,therecanbeC(10,6)*C(15,5)teams

　　Q2:

　　(i)AsSeventypercentofthefirst-yearstudentsaremale,thereare

　　12(40*30%)studentsarefemalefrom40first-yearstudents.andsevenofthefemalestudentsgraduatedfromhighschooltheyearbefore,so5(12-7)femalestudentstooka“gap-year”；

　　Toconclude,theprobabilityis1/8（5/40）thatarandomlyselectedfirst-yearstudentisbothfemaleandtooka“gap-year”.

　　（ii）Theprobabilityis1/2（halfof40）thatgivenarandomlyselectedfrom40first-yearstudentsare“gap-year”ones,andtheprobabilityis3/8[（1/2）*（15/20）]thatgivenarandomlyselected40first-yearstudesaremalefrom“gap-year”ones.

　　So,theprobabilityis3/4[（3/8）/（1/2）]thatgivenarandomlyselectedstudenttooka“gap-year”thestudentismale.

　　Q3：

　　(i)thereare22（13+12-3）redcardsorfacecards,sotheprobabilityis11/27（22/54）thatthecarddrawnisaredcardorafacecard.

　　（ii）theprobabilityofhavingonlyoddnumbersresultwhenanunbiasedsix-sideddieisrolledonetimeis1/2（3/6）；

　　whenanunbiasedsix-sideddieisrolledfivetimes,theprobabilityofhavingonlyoddnumbersresultis1/32（1/2*1/2*1/2*1/2*1/2）.