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M=male, F=female, S=student, P=probability. 20 students=12F+8M.

PROBABILITY OF SELECTING JUST 1F

If we select F as S1 (S1=F, first student) then we have 19S=11F+8M remaining. P=12/20 for selecting F as S1.

The remaining 7 in the random selection must be M, so the probability changes as fewer S remain.

S2=M, P=8/19, 18S=11F+7M; S3=M, P=7/18, 17S=11F+6M; S4=M, P=6/17, 16S=11F+5M; S5=M, P=5/16, 15S=11F+4M; S6=M, P=4/15, 14S=11F+3M; S7=M, P=3/14, 13S=11F+2M; S8=M, P=2/13, 12S=11F+1M.

However, any of the selected S could be F provided the rest were M, so there are 8 positions for F in the selection. The probability is therefore 8(12/20)(8/19)(7/18)(6/17)(5/16)(4/15)(3/14)(2/13)=0.000762 approx.

PROBABILITY OF SELECTING JUST 1M

P=8(8/20)(12/19)(11/18)(10/17)(9/16)(8/15)(7/14)(6/13)=0.050298

PROBABILITY OF SELECTING JUST 2F

The number of ways 2F can appear within 8S is 28.

P=28(12/20)(11/19)(8/18)(7/17)(6/16)(5/15)(4/14)(3/13)=0.014670

PROBABILITY OF SELECTING JUST 3M

The number of ways 3M can appear within 8S is 56.

P=56(8/20)(7/19)(6/18)(12/17)(11/16)(10/15)(9/14)(8/13)=0.352084

by Top Rated User (982k points)

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