Question #296726

A family has four children. If X is a random variable that pertains to the number of female children, what are the possible values of X?


1
Expert's answer
2022-02-14T10:29:23-0500

The possible values of XX are: 0, 1, 2, 3, 4


P(X=0)=P(BBBB)=(12)4=116P(X=0)=P(BBBB)=(\dfrac{1}{2})^4=\dfrac{1}{16}

P(X=1)=P(GBBB)+P(BGBB)P(X=1)=P(GBBB)+P(BGBB)

+P(BBGB)+P(BBBG)=4(12)4=14+P(BBGB)+P(BBBG)=4(\dfrac{1}{2})^4=\dfrac{1}{4}

P(X=2)=P(GGBB)+P(GBBG)P(X=2)=P(GGBB)+P(GBBG)

+P(GBGB)+P(BGGB)+P(BGBG)+P(GBGB)+P(BGGB)+P(BGBG)

+P(BBGG)=6(12)4=38+P(BBGG)=6(\dfrac{1}{2})^4=\dfrac{3}{8}

P(X=3)=P(GGGB)+P(GGBG)P(X=3)=P(GGGB)+P(GGBG)

+P(GBGG)+P(BGGG)=4(12)4=14+P(GBGG)+P(BGGG)=4(\dfrac{1}{2})^4=\dfrac{1}{4}



P(X=4)=P(GGGG)=(12)4=116P(X=4)=P(GGGG)=(\dfrac{1}{2})^4=\dfrac{1}{16}

x01234p(x)1/161/43/81/41/16\def\arraystretch{1.5} \begin{array}{c:c} x & 0 & 1 & 2 & 3 & 4 \\ \hline p(x) & 1/16 & 1/4 & 3/8 & 1/4 & 1/16 \\ \end{array}


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