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$$P(R)=1/5$$ For each forecast $i=1,2,3$ $$P(H_i = R|R)= 3/4$$ and $$P(H_i=R|\bar{R})=1/4$$ We need to find $$P(R|H_1=R, H_2= R, H_3 = \bar{R}) = \frac{P(R) P(H_1=R, H_2=R, H_3=\bar{R}|R)}{P(H_1=R, H_2=R, H_3=R)} = \frac{P(R) P(H_1=R, H_2=R, H_3=\bar{R}|R)}{P(R) P(H_1=R, H_2=R, H_3=\bar{R}|R)+P(\bar{R}) P(H_1=R, H_2=R, H_3=\bar{R}|\bar{R})} =\frac{1/5 ... 1 Using Bayes' rule, for each value x you have$$P(x | data) \propto P(data | x) prior(x)$$Therefore, P(x|data) is zero if x \neq 0 and so P(0|data) = 1, so your posterior is the same as the prior. No, the posterior is the same as the prior. Using this prior says that you have decided that the answer is zero and no evidence can convince you ... 0 By using Bayes' Theorem, what I am getting is that$$P(\mathbf{Urn1} | RR) = \frac{(P(RR|\mathbf{Urn1})) * P(\mathbf{Urn1})}{P(RR)}$$Where P(RR|\mathbf{Urn1}) = 9/16  and P(\mathbf{Urn1}) = 1/2. To find P(RR) we use the law of total probability:$$P(RR) =P(RR|\mathbf{Urn1}) *P(\mathbf{Urn1}) + P(RR|\mathbf{Urn2}) * P(\mathbf{Urn2})$$We know that ... 2 Just hints, as requested: Let U_i be the event that you are drawing from urn i and let R_i be the event that you draw a red ball on your ith draw (with replacement). Here are some questions to guide your thinking: If you know you are drawing from a particular urn, with replacement, what is the applicable probability model for drawing red balls? ... 0 You do the same reasoning, but you replace your equation :$$p(D=0) = \sum_{G\in\{0,1\}}(p(D=0\mid G)\sum_{F,B\in\{0,1\}}(p(G\mid F,B)p(F)p(B))) = 0,352$$by$$p(D=0 | F=0) = \sum_{G\in\{0,1\}}(p(D=0\mid G)\sum_{B\in\{0,1\}}(p(G\mid F=0,B)p(B))) $$Anyway, for this kind of problem, it should be easier to use the Pearl's inference algorithm that uses ... 0 Consider what happens with testing a million people (having the right proportion of sick people): 100 of them have "the giggles" and if we test them we find 99 positive tests, as the test is 99% reliable. Of the remaining (1000000 - 100) = 999900 non-"giggles" people and 1% of them test positive (as 99% of them test negative), which means 9999 positive ... 0 Hint: Bayes' Rule helps keep track of the bookkeeping involved when using conditional probabilities. Try breaking down your question: What events are we talking about. Let G be "have the giggles" and T^+ be a positive test result. We know P(G|T^+) and P(\neg G|\neg T^+), right? We also know that P(G)=1/10000. Do you know Bayes' Rule? Can you ... 1 For an introduction, I would recommend Probabilistic Programming & Bayesian Methods for Hackers by Cam Davidson-Pilon, freely available online. http://camdavidsonpilon.github.io/Probabilistic-Programming-and-Bayesian-Methods-for-Hackers/ 1 Writing D|X=x \sim N(0, 14^2) for all values of x emphasize that the random variable D is independent of X. Of course in this case we know that D and D|X = x has the same distribution as indicated by independence. Maybe, depending on the context, the distribution of interest is this conditional distribution so the author will need to present this ... 0 The integral is easily calculated using the integral representation of the Dirac delta$$ \delta(x)=\int_{-\infty}^\infty\frac{dk}{2\pi}e^{\mathrm{i}kx}\ . $$We get$$ \int_{-\infty}^{\infty}...\int_{-\infty}^{\infty} \delta_\mathrm{D}\left(\sum_{i=1}^M f_i - 1\right)\prod_{i=1}^M \Theta(f_i) \,\mathrm{d}f_i = ...

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No.   You cannot infer conditional independence from pairwise independence. Applying Bayes' Rule can only obtain: \begin{align} \mathsf P(A, B\mid D) ~=~& \dfrac{\mathsf P(D\mid A, B)~\mathsf P(A,B)}{\mathsf P(D)} \\[1ex] =~& \dfrac{\mathsf P(D\mid A, B)~\mathsf P(A)~\mathsf P(B)}{\mathsf P(D)} \end{align}

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Taking a Bayesian approach one considers the results on the first $n$ tosses to provide information about the Heads probability $\theta.$ So if you got $r = 400$ heads in $n = 1000$ previous tosses, your posterior distribution would be $\theta \sim Beta(401, 601).$ Taking the posterior mean as the success probability for toss $1001$ you would have $E(\theta) ... 1 We do not need to use Bayes' theorem. By the law of total probability, \begin{multline*} P(\text{correct bit})=P(\text{correct bit}\mid\text{$1$was sent})P(\text{$1$was sent})\\+P(\text{correct bit}\mid \text{$0$was sent})P(\text{$0$was sent}). \end{multline*} Hence, $$P(\text{correct bit})=0.98\cdot0.5+0.99\cdot0.5=0.985.$$ 3 Everything is correct, except your intuition, which is only partially right:$E[X|Y=0]=0$, not$1/2$. In fact, the smaller$Y$is, the more likely$X$is to be small. More precisely, the conditional density of$X$at$x$given$Y=y$is proportional (by Bayes) to that of$Y$at$y$given$X=x$, that is, to$1/x$. And since$1/x$integrates to infinity on ... 0 Let's say your rating system says A consistently has a$60\%$chance of winning against B. You have the following match data available: in four matches, the victors were A, A, B, A. In reality, the results of matches are not independent of each other (contestant confidence and other factors are in effect) but let's assume they are. The prior odds of A ... 0 The 1997 paper here by Julier and Uhlmann seems a good place to start. Their assumed state update model is slightly more general than yours, being the vector time varying model $$x(k+1)=f_k[x(k),u(k),w(k)].$$ You can also check the more recent literature citing this paper. A technique they utilize in 1 seems to be what's called the "unscented kalman ... 0 Wait, why do you want to avoid getting the mean value if zero questions are answered? That corresponds to the roughest possible estimate, which is what you should get if no information is given (zero questions are answered). Think of conditional expectation: conditioning on zero questions answered is roughly equivalent to conditioning on the trivial sigma ... 2 There are some correct answers here. Many use Bayes' rule, which is correct and elegant but takes getting used to. Let me try instead to help you think through this particular example, to train your intuition. In your answer to #1 you correctly compute that the probability of one$T$is 1/2. But that doesn't mean the probability of$TTT$is 1/8 unless you ... 2 I will extend my comment: remember that$\text{probability of A}=\frac{\text{number of A cases}}{\text{all possible cases}}\$. Then, how many ways we can get (tail, tail, tail)? If we take the fair coin with this coin we only can take (tail, tail, tail) i.e. only exist one way we can take the desired result. But if we took the double-tail coin we take ...

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By Bayes' theorem \begin{align}P(M_3\mid TTT)=\frac{P(TTT\mid M_3)P(M_3)}{P(TTT)}=\frac{\left(\frac12\right)^3\cdot\frac13}{\frac38}=\frac19\end{align} Note: The denominator was calculated using the Law of total probability as is common when applying the Bayes rule. You did this in part 1. but not correctly. To see this write ...

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Paint the double-head coin yellow on one side and red on the other. Paint the double-tail coin blue and green. There are 24 possible outcomes. One of the outcomes is: Pick double-head, toss yellow,yellow,yellow Run through the 24 outcomes, how many of them give you three tails? Of those outcomes, how many were with the fair coin, how many were with ...

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