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Let $X$ denote the total number of games played, and let $J$ denote the number of games Joe played. The conditional distribution of $X$ given $J=4$, namely $X|J=4$, is supported on $\{4,5,\ldots\}$ and has pmf $$P(X=k|J=4)=\frac{P(J=4|X=k)P(X=k)}{\sum_{k=4}^{\infty}P(J=4|X=k)P(X=k)}$$ which is non zero whenever $k\geq 4$. Using the facts that $X\sim \text{... 3 Given that the random vector$\textbf X$has certain number of entries$n$, it follows the multinomial distribution parameters$n, \textbf p$. This is because$X$is created by incrementing each coordinate with the vector of probabilities$\textbf p$,$ntimes. Then the joint distribution is found by: \begin{split}f(x_1,...,x_m,z)&=f(x_1,...,x_m|z)\Pr(... 3 I'd like to obtain: number of Electric Vehicles (EVs) which arrive to a charging station during one day; and their Time-of-Arrivals (ToA); Unfortunately the total number of arrivals in a day is not deterministic quantity thus you cannot "know" it but you can have a probability information about how many they are. It is a random variable, ... 3 \begin{align} & \Pr(S_T\le s) = \operatorname E(\Pr(S_T\le s\mid T)) \\[8pt] = {} & \sum_{t=1}^\infty \Pr(S_t\le s\mid T=t) \Pr(T=t) \\[8pt] = {} & \sum_{t=1}^\infty \int_0^s \frac 1 {\Gamma(t)} (\lambda u)^{t-1} e^{-\lambda u} (\lambda\, du) \cdot (1-p)^{t-1} p \\[8pt] = {} & \int_0^s \sum_{t=1}^\infty \frac 1 {(t-1)!}(\lambda u(1-p))^{t-1} ... 2 Your approach is fine, though the sum should begin at k=1 (or k=2). I don't think the expression has a nice closed form. Noticing that the mean of the variable is \lambda = 10, and that the approximation (1-1/n)^n \approx e^{-1} should work reasonably around n\approx 10, then the function g(n)= n \, (1-1/n)^n is approximately linear around the ... 2 The number of virus attacks are dependent on the time that the PC is on, thus the distribution of the virus attacks isV|T\sim Po\left(\frac{t}{5}\right)$$and thus$$f_{VT}(v,t)=\frac{e^{-t/5}\left(\frac{t}{5}\right)^v}{v!}\cdot\frac{5}{4t^2}$$The probability that V=2 is$$\frac{1}{40}\int_1^5 e^{-t/5}dt=\frac{5e^{-1/5}-5e^{-1}}{40}\approx 5.64\%$$... 2 For \phi \in [0,10] the log likelihood is:$$\log\left[ \frac{\exp(-\phi) \phi^2}{2!} \frac{\exp(-2\phi) (2\phi)^4}{4!} \right]=-3\phi+6 \log\phi +\log16-\log(2!)-\log(4!)$$Setting the derivative to 0 gives -3+6/\phi=0 or \phi=2. 2 If phone calls arrive at a switchboard at an average rate of 3 per minute, and at another switchboard at an average rate of 4.2 per minute, then at the two switchboards combined, they arrive at an average rate of 7.2 per minute. 2 Number in a day is modeled by \text{Poisson}\left(rt\right). Arrival times are modeled by \text{Gamma}(n, r) with n being the nth arrival. 2 While Matthew Pilling and tommik provide answers that show the mathematics of getting the answer, I will provide the intuition involved. We know that, on this specific day, Joe played 4 solos. This provides a minimum number of songs - specifically, there must have been at least 4 songs. Note that it is very possible for the Poisson distribution to produce 0 ... 2 The answer of @Matthew Pilling is perfect (+1). A Bayesian approach will lead to the same solution avoiding a lot of calculations: (constants are not considered until the end of the process)$$\mathbb{P}[X|J=4]\propto \mathbb{P}[X]\cdot \mathbb{P}[J=4|X]\propto\frac{5^x}{x!}\cdot \binom{x}{4}\left(\frac{1}{2}\right)^x\propto\frac{\left(\frac{5}{2}\right)^{x}}... 1 DefineX$to be the payment by the company in one month. The number of accidents in one month is distributed$\text{Poisson}(.3)$. Define a variable$Y$that is the number of accidents per month, then the event that$X=5000$is the same as$Y=0$and$X=10000$is the same as$Y=0\text { or } 1$. So using the definition of expected value $$\begin{split}E(X)&... 1 For b, you want the conditional probability$$P(X=8|X\geq 6) = \frac{P(X=8 \cap X \geq 6)}{P(X\geq 6)}= \frac{P(X=8)}{P(X\geq 6)}$$For c, you want to find the probability of zero calls between 12:30 and 13:00 on any day, and then use a binomial distribution. You have 7 days, and you want any 4 of them to have zero. Remember, the requirements for a ... 1 Let X be the earnings, then by definition of expected value,$$\begin{split}\mathbb E(X)&=60 \mathbb P(X=60)+70\mathbb P(X=70) +80\mathbb P(X=80)\end{split}$$Let p=.0803, the chance that there are 3 or more defects on the roll. The inspector earns \60 if he inspects 4, 5, or 6 total rolls, given by probability p^4+4p^4(1-p)+{5\choose 2}p^4(1-... 1 To get an answer it is necessary to fix a certain n. So let's set n=5 as per Neyman Pearson's Lemma, the critical region is$$\mathbb{P}[Y\geq k]=0.05$$where Y\sim Po(5) It is easy to verify with a calculator (or manually in 5 minutes) that$$\mathbb{P}[Y\geq 10]=3.18\%$$and$$\mathbb{P}[Y\geq 9]=6.81\%$$It is evident that there's no way to have a ... 1 The answers are the same, notice 1-e^{-\lambda_1t}+\frac{\lambda_1}{\lambda_1+\lambda_2}e^{-\lambda_1 t}=1-\frac{\lambda_2}{\lambda_1+\lambda_2}e^{-\lambda_1t}. For your calculation the \lambda_2 in the second term was somehow dropped$$\begin{split}\Pr(T_1<T_2+t)&=\int_0^\infty\Pr(T_1<T_2+t|T_2=t_2)f_{T_2}(t_2)dt_2\\ &=\int_0^\infty\left(1-... 1 Just note that for$a,b\in \Bbb Z\begin{align*} F_{X,W}(a,b)&=\Pr [X\leqslant a, X+Y\leqslant b]\\ &=\sum_{\{(t,s)\in \mathbb{Z}^2:t\leqslant a,t+s\leqslant b\}}f_{X,Y}(t,s)\\ &=\sum_{(t,s)\in\mathbb{Z}^2}\mathbf{1}_{(-\infty ,a]}(t)\mathbf{1}_{(-\infty ,b]}(t+s)f_{X,Y }(t,s)\\ &=\sum_{(t,s)\in\mathbb{Z}^2}\mathbf{1}_{(-\infty ,a]}(t)\... 1 If by "rate \lambda" you mean a process X_t such that \mathbb E[X_t] = \lambda t, then this is just linearity of expectation:\mathbb E[2 X_t + 3 Y_t] = 2 \mathbb E[X_t] + 3 \mathbb E[Y_t] = (2\lambda + 3 \mu) t$$On the other hand, 2 X_t + 3 Y_t is certainly not a Poisson process (e.g. this never takes the value 1), while if X_t and ... 1 Simple answer is that for both problems, you have a span of two hours, so the calculation is identical for both of them. On average 7 customers arrive in 1 hour so 14 customers arrive in 2 hours. Using X\sim\text{Poisson}(14) to be the number of customer arrivals in a 2 hour time interval, we have$$\Pr(X=2)=\frac{e^{-14}(14)^2}{2!}$$In terms of a Poisson ... 1 You can try doing it directly. Let f(\lambda) be the log-likelihood after ignoring the additive terms that do not involve \lambda, then (please check my calculation)$$ f(\lambda) = -n\lambda+\ln{\lambda}\sum {x_i} +n\ln{\lambda} -\lambda\sum{y_i}. $$Then differentiating with respect to \lambda and equating it to zero get$$ \hat{\lambda}= \frac{1+\... 1 Because the application of the poisson distribution is an approximation only. The quality of a tablet is binomial distributed asX\sim \textrm{Bin}(500,0.001)$. Then $$P(X=0)=\binom{500}{0}\cdot 0.01^0\cdot 0.99^{500}=0.00657...=0.6750\%$$ But your approximation is not so bad. 1 First, the mean of$X_1-Y_1$is$\lambda-\lambda^{-1}$. Thus, the MM estimator of$\lambda$is $$\hat{\lambda}_n=\frac{1}{2}\left(m_n+\sqrt{m_n^2+4}\right),$$ where$m_n:=n^{-1}\sum_{i=1}^{n}(X_i-Y_i)$. For the asymptotic distribution of$\hat{\lambda}_n\$, note that $$\sqrt{n}\left(m_n-(\lambda-\lambda^{-1})\right)\xrightarrow{d}N\!\left(0,\lambda+\lambda^... 1 A Poisson Distribution gives the probability of a number of events in an interval generated by a Poisson process. The Poisson distribution is defined by the rate parameter, \lambda, which is the expected number of events in the interval: expected here means "on average". It's all about the average number of events in the given time interval. Now ... 1 You have an error in your exponential. Its e^{-\lambda t}. Then you use conditional probabilities and denote the poisson process by (N_t)_{t\geq 0}:$$P(X_t = k)=\sum_{n=0}^{\infty} P(X_t = k| N_t = n)P(N_t = n)=\sum_{n=0}^{\infty} \left(\dfrac{1}{6}\right)^n P(N_t = n)=\sum_{n=0}^{\infty} \left(\dfrac{1}{6}\right)^n \dfrac{e^{-\lambda t}(\lambda t)^n}{n!...