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They will have either 1 or 0 girls. Let $y$ be the probability that a child is a girl, then $E(Y) = 1-(1-y)^5$. $E(X) = 5 (1-y)^4 + 4 (1-y)^3y + 3 (1-y)^2y + 2 (1-y)y + y$

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The correlation coefficient measures the strength of linear relation between two variables. The closer the correlation is to plus-one or minus-one the stronger the linear relationship.   A correlation coefficient of exactly plus-one means there is a perfect, direct, increasing linear-relation.   A correlation coefficient of exactly minus-one means ...

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\begin{eqnarray} P [ X+Y \le \alpha] &=&\int_{x=-\infty}^\infty \int_{y=-\infty}^\infty 1_{\{(x,y) | x+y \le \alpha \}} ((x,y)) f_X(x) f_Y(y)dy dx \\ &=& \int_{x=-\infty}^\infty \int_{y=-\infty}^{\alpha-x} f_X(x) f_Y(y) dy dx \\ &=& \int_{x=-\infty}^\infty \int_{y=-\infty}^{\alpha} f_X(x) f_Y(y-x) dy dx \\ &=& ...

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I am not quite sure, but I think your error is $P(X \le x, Y \le t-X) = P(X \le x, Y \le t-x)$. The most you can say here is "$\le$"; I don't think equality holds. It suffices to show $f_{X,T}(x,t) = f_X(x) f_Y(t-x)$; then, integrating over $x$ gives the desired form of $f_T(t)$. My computation below is a little unrigorous because I condition on an event ...

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I have reproduced what your friend has done. You are right that it is plain wrong. In the image I have given a counterexample of the choice of b values for 50% correct scenario and have illustrated to you that that correlation is not 0 depending on the choice of the values that your b takes for those that do not move the same way. To be polite, the ...

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Both "Cov" and "Var" are used to represent the covariance matrix of the vector $\bf X$. See for example this Wikipedia remark: Nomenclatures differ. Some statisticians, following the probabilist William Feller, call the matrix $\Sigma$ the variance of the random vector X, because it is the natural generalization to higher dimensions of the 1-dimensional ...

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Call the one particular black ball we're concerned with $B$. Then the orders in question are:  \begin{array}{cccccccccl} W & W & W & \cdots & W & W & W & B & & w!\text{ different orders} \\ W & W & W & \cdots & W & W & B & W & & w!\text{ different orders} \\ W & W & W & ...

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I have a heuristic, very intuitive idea, so this is not a proof, just some thinking direction. Short: we possibly can take the conditional expectation over all the subsets of subalgebra $\sigma(C) \subset \mathcal{C}$, where $\mathbb{E}[X \mathbb{1}_{C}]\geq 0, \ C \in \mathcal{C}$ and set $Y=0$ elsewhere. At first, it is obvious that if $X$ takes only ...

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