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Solution
a)According to given probelm total students=100(28+6+8+8+6+44)
the probability that a student selected at random from the fresman class is a female is
P(F)=(28/100)*(0.32)+(6/100)*(0.61)+(8/100)*(0.62)+(8/100)*(0.51)+(6/100)*(0.57)+(44/100)*(0.47)
=(0.28)*(0.32)+(0.06)*(0.61)+(0.08)*(0.62)+(0.08)*(0.51)+(0.06)*(0.57)+(0.44)*(0.47)
=0.0896+0.0366+0.0496+0.0408+0.0342+0.2068
=0.4576
b)The probability that a business student selected at random from the freshman class is male is
P(B/M)=[P(B)P(M/B)]/P(M)
now P(M)=(28/100)*(0.68)+(6/100)*(0.39)+(8/100)*(0.38)+(8/100)*(0.49)+(6/100)*(0.43)+(44/100)*(0.53)
=(0.28)*(0.68)+(0.06)*(0.39)+(0.08)*(0.38)+(0.08)*(0.49)+(0.06)*(0.43)+(0.44)*(0.53)
=0.1904+0.0234+0.0304+0.0392+0.0258+0.2332
=0.5424
then P(B/M)=[P(B)P(M/B)]/P(M)
=[(28/100)*(0.68)]/[(28/100)*(0.68)+(6/100)*(0.39)+(8/100)*(0.38)+(8/100)*(0.49)+(6/100)*(0.43)+ (44/100)*(0.53)]
=[(0.28)*(0.68)]/[0.5424]
=(0.1904)/(0.5424)
=0.3510
c)The probability that a female student selected at random from the freshman class in majoring in business is
P(B/F)=[P(B)P(F/B)]/P(F)
                          =[(28/100)*(0.32)]/[0.4576]      [from a]
                          =[(0.28)*(0.32)]/[0.4576]
=[0.0896]/[0.4576]
=0.1958

