Answer to Question #257974 in Statistics and Probability for Shaon

Question #257974

1.    Ajouin and Genie are a happy couple and have 4 lovely children. Is it more likely they will have two boys and two girls or three of the same sex and one of the other? Assume that the probability of a child being a boy is 0.5 and that the births are independent events. (5 points)

2.    A fair die is rolled 4 times. Let the random variable X denote the number of sixes that appear. Find FX(x). (5 points)


1
Expert's answer
2021-11-17T15:33:22-0500

1.



"P(bbbb)=(\\dfrac{1}{2})^4=\\dfrac{1}{16}""P(bbbg)=P(bbgb)=P(bgbb)=P(gbbb)""=(\\dfrac{1}{2})^4=\\dfrac{1}{16}""P(bbgg)=P(bgbg)=P(bggb)=P(gbbg)=P(gbgb)=P(ggbb)""=(\\dfrac{1}{2})^4=\\dfrac{1}{16}""P(gggb)=P(ggbg)=P(gbgg)=P(bggg)""=(\\dfrac{1}{2})^4=\\dfrac{1}{16}""P(gggg)=(\\dfrac{1}{2})^4=\\dfrac{1}{16}""P(2\\ boys\\ \\&\\ 2\\ girls)=6(\\dfrac{1}{16})=\\dfrac{3}{8}""P(3\\ boys \\ \\&\\ 1\\ girl)+P(1\\ boy \\ \\&\\ 3\\ girls)""=4(\\dfrac{1}{16})+4(\\dfrac{1}{16})=\\dfrac{1}{2}>\\dfrac{3}{8}"

It is more likely they will have three of the same sex and one of the other than two boys and two girls.


2. "X\\sim Bin(n, p)"

"n=4, p=\\dfrac{1}{6}, q=1-p=\\dfrac{5}{6}"



"P(X=0)=\\dbinom{4}{0}(\\dfrac{1}{6})^0(\\dfrac{5}{6})^{4-0}=\\dfrac{625}{1296}""P(X=1)=\\dbinom{4}{1}(\\dfrac{1}{6})^1(\\dfrac{5}{6})^{4-1}=\\dfrac{500}{1296}""P(X=2)=\\dbinom{4}{2}(\\dfrac{1}{6})^2(\\dfrac{5}{6})^{4-2}=\\dfrac{150}{1296}""P(X=3)=\\dbinom{4}{3}(\\dfrac{1}{6})^3(\\dfrac{5}{6})^{4-3}=\\dfrac{20}{1296}""P(X=4)=\\dbinom{4}{4}(\\dfrac{1}{6})^4(\\dfrac{5}{6})^{4-4}=\\dfrac{1}{1296}""F_X(x) = \\begin{cases}\n 0, &x<0 \\\\\n 625\/1296, &0\\leq x<1\\\\\n 1125\/1296, &1\\leq x<2\\\\\n 1275\/1296, &2\\leq x<3\\\\\n 1295\/1296, &3\\leq x<4\\\\\n 1, &x\\geq 4.\n\\end{cases}"

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