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3

\begin{eqnarray*} P\left(F\right) &=& P\left(F\vert H\right)P\left(H\right) + P\left(F\vert L\right)P\left(L\right) \\ &=& P\left(F\vert H\right)P\left(H\right) + \left(1 - P\left(U\vert L\right)\right)P\left(L\right) \\ &=& 0.9 \times 0.3 + \left(1 - 0.6\right) \times 0.7 \\ &=& 0.27 + 0.28 \\ &=& 0.55 \end{eqnarray*} ...

3

#of possibilities in sample space = #tickets in A * #tickets in B (here $4\cdot 3=12$) #of possibilities meet the constraint: Sum over thee cases of chosen tickets in Box : for (x,6) all 4 tickets in Box A meets the conditions. For (x,5) on less -> 3, for (x,4) one more less ->2. The sum is 4+3+2=9. Hope this Hint can help you to generalize your solution ...

2

Factoid: For every $(u,v)$, if $X$ is standard normal, then $$E(\phi(uX+v))=\int\phi(ux+v)\phi(x)\mathrm dx.$$ Simple algebraic manipulations show that $$\phi(ux+v)\phi(x)=\phi(uvw+w^{-1}x)\phi(vw),\qquad \color{red}{w=(u^2+1)^{-1/2}},$$ hence, using the change of variable $z=uvw+w^{-1}x$, one gets the key-identity ...

2

Re 1., note that the random variable $X=\limsup\limits_{t\to\infty}B_t/\sqrt{t}$ is asymptotic hence, by Kolmogorov zero-one law, if $P(X\gt0)\lt1$, then $P(X\gt0)=0$, that is, $X\leqslant0$ almost surely. If this holds, then, by symmetry, $\liminf\limits_{t\to\infty}B_t/\sqrt{t}\geqslant0$ almost surely, thus, $B_t/\sqrt{t}\to0$ almost surely. In ...

1

Ok, you've accepted an answer. Here's a generalization for consecutive integers in the two sets. $A=\{n,\ldots,n+m\}$, $\quad B=\{k,\ldots,k+l\}$, $n,k>0$ and $m,l\geq0$. Let's say you want them to sum to the greater than the number $d \in \Bbb N$. Suppose you know the least $a \in \Bbb N,\; 0\leq a \leq m$ s.t. there exists a \$b \in \Bbb N, \; 0\leq b ...

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