A family has four children. If Y is random variable that pertains to the number of female children what are the possible values of Y?
This is a binomial distribution with n=4,p=12.n=4,p=\frac{1}{2}.n=4,p=21.
P(Y=y)=C42(12)x(12)4−x.P(Y=y)=C_4^2(\frac{1}{2})^x(\frac{1}{2})^{4-x}.P(Y=y)=C42(21)x(21)4−x.
P(Y=0)=C40(12)0(12)4=116.P(Y=0)=C_4^0(\frac{1}{2})^0(\frac{1}{2})^4=\frac{1}{16}.P(Y=0)=C40(21)0(21)4=161.
P(Y=1)=C41(12)1(12)3=14.P(Y=1)=C_4^1(\frac{1}{2})^1(\frac{1}{2})^3=\frac{1}{4}.P(Y=1)=C41(21)1(21)3=41.
P(Y=2)=C42(12)2(12)2=38.P(Y=2)=C_4^2(\frac{1}{2})^2(\frac{1}{2})^2=\frac{3}{8}.P(Y=2)=C42(21)2(21)2=83.
P(Y=3)=C43(12)3(12)1=14.P(Y=3)=C_4^3(\frac{1}{2})^3(\frac{1}{2})^1=\frac{1}{4}.P(Y=3)=C43(21)3(21)1=41.
P(Y=4)=C44(12)4(12)0=116.P(Y=4)=C_4^4(\frac{1}{2})^4(\frac{1}{2})^0=\frac{1}{16}.P(Y=4)=C44(21)4(21)0=161.
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