Answer to Question #305475 in Statistics and Probability for BELLAFAYRE

Question #305475

The number of females employees selected if five employees are selected by lottery from a group of six male and six female employees.

1
Expert's answer
2022-03-04T05:19:13-0500

Let X=X= the number of females employees selected. The possible values of the random variable XX are 0,1,2,3,4,5.0, 1, 2,3,4,5.

There are (6+65)=792\dbinom{6+6}{5}=792


P(X=0)=(60)(650)(125)P(X=0)=\dfrac{\dbinom{6}{0}\dbinom{6}{5-0}}{\dbinom{12}{5}}

=1(6)792=1132=\dfrac{1(6)}{792}=\dfrac{1}{132}


P(X=1)=(61)(651)(125)P(X=1)=\dfrac{\dbinom{6}{1}\dbinom{6}{5-1}}{\dbinom{12}{5}}

=6(15)792=15132=\dfrac{6(15)}{792}=\dfrac{15}{132}



P(X=2)=(62)(652)(125)P(X=2)=\dfrac{\dbinom{6}{2}\dbinom{6}{5-2}}{\dbinom{12}{5}}

=15(20)792=50132=\dfrac{15(20)}{792}=\dfrac{50}{132}



P(X=3)=(63)(653)(125)P(X=3)=\dfrac{\dbinom{6}{3}\dbinom{6}{5-3}}{\dbinom{12}{5}}

=20(15)792=50132=\dfrac{20(15)}{792}=\dfrac{50}{132}





P(X=4)=(64)(654)(125)P(X=4)=\dfrac{\dbinom{6}{4}\dbinom{6}{5-4}}{\dbinom{12}{5}}

=15(6)792=15132=\dfrac{15(6)}{792}=\dfrac{15}{132}





P(X=5)=(65)(655)(125)P(X=5)=\dfrac{\dbinom{6}{5}\dbinom{6}{5-5}}{\dbinom{12}{5}}

=6(1)792=1132=\dfrac{6(1)}{792}=\dfrac{1}{132}





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