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3

Setup on recursion: $p\left(n,i\right)$ probability that $i$ of $n$ candies are eaten. To be found: $p\left(20,k\right)$. $p\left(n,i\right)=0$ if $2i<n\vee i>n$ $p\left(0,0\right)=1=p\left(1,1\right)$ If $2i\geq n\geq2$ then: $$p\left(n,i\right)=0.04\left(n-1\right)p\left(n-2,i-1\right)+\left[1-0.04\left(n-1\right)\right]p\left(n-1,i-1\right)$$ ...


1

Yes. The revelation principle states that if there is a social choice function that can be implemented by some mechanism (i.e., if that mechanism has an equilibrium outcome that corresponds to the outcome of the social choice function), then there is a direct mechanism that also implements that choice function. Hence, if there exists an auction (mechanism) ...


5

You're bringing causation and temporality in where they don't belong. $A\cap B$ is merely the event which occurs if both $A$ and $B$ occur. There's no implication of temporal order or cause and effect. Events are sets, and $\cap$ has the usual set-theoretic meaning here; hence it's symmetric.


0

$$\begin{align}\mathsf P(R) =&~ 0.1\\ \mathsf P(PA) =&~ 0.5\\ \mathsf P(PR\mid R, PA) =&~ 0.6\\ \mathsf P(PR\mid ¬R, PA) =&~ 0.4\\\mathsf P(PR\mid ¬R, ¬PA) =&~ 0.1\\ \mathsf P(PR\mid R, ¬PA) =&~ 0.2\end{align}$$ What is the probability of $\mathsf P(¬R, PR, ¬PA)$? $$\begin{align}\mathsf P(\neg R, PR, \neg PA)=&~\mathsf P(PR\...


2

The probability of catching exactly one fish are shaded below in Venn diagrams for each pond separately, where the three circles for a given pond correspond to catching a fish on the first, second, and third tries, respectively. (Areas not to scale.) For Pond B, where the calculation is partly shown, the probability of catching exactly one fish is $3\cdot\...


1

You can supply some of the missing information through plausible assumptions explicitly stated and the get answers conditioning on those assumptions that will apply or not depending on the verifiability of those assumptions. For instance, you can explicit the assumption that, given the chosen lake, each cast will be a realization of an independent Bernouli ...


0

Let $A_i$ be the event were our ith picking yields a working element. Also let $S$ be the chosen series, with values $a, b, c$, for the series with $15, 18, 16$ working elements, respectively. Then we can see that: $$P(A_i|_{S=a}) = \frac{15}{20}, P(A_i|_{S=b}) = \frac{18}{20}, P(A_i|_{S=c}) = \frac{16}{20}$$ With those values we can obtain the probability ...


1

Let us define some events in order to clean up notation a bit. Let $W_1,W_2$ be the events that the first element works and second element works respectively. Let $S_1,S_2,S_3$ be the events that we are pulling items from series $1,2$ and $3$ respectively. The problem asks us "What is the probability that a second element drawn from the same series as the ...


0

First, observe that $$\prod_{i=1}^n \lambda e^{-\lambda x_i} = \lambda^n e^{-\lambda \sum x_i} = \lambda^n e^{-n \bar x \lambda},$$ where $\bar x = \frac{1}{n} \sum_{i=1}^n x_i$ is the sample mean. Then remove every factor that is not a function of $\lambda$: $$f(\lambda \mid \boldsymbol x) \propto \lambda^n e^{-n \bar x \lambda} \lambda^{\alpha - 1} e^{-\...


2

This answer is based on the book "Optimal and Robust Estimation" by Lewis, Xie and Popa. Let's say you have this a system with the dynamics: $$x_{k+1}=A_kx_k+B_ku_k+G_kw_k$$ Where $u_k$ is the input and $w_k$ is a Gaussian noise. Also, let the observation equation be: $$z_k=H_kx_k+v_k$$ Now again, $v_k$ is a Gaussian noise process. Your prior for the states ...


0

$\textbf{Hint: }$ If $R$ is at $1$ or $20$ at time $t$, can $H$ win at time $t+1$?



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