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## Hot answers tagged probability

3

Yes so when you roll one dice and you would like to obtain a $2$, the chance is $\frac{1}{6}$, when there are two die, there are now $36$ possible outcomes which you can list for example $(1,1) (1,2)..... (6,6)$. For this there are $11$ outcomes, try and write out yourself it will make more sense! So the chance is $\frac{11}{36}$ which is greater than only ...

2

We have $$P(X = k | X+Y = m) = \frac{P(X = k \,\&\, Y = m-k)}{P(X+Y = m)} = \frac{p_{k}p_{m-k}}{\sum_{j=0}^{m}p_{j}p_{m-j}} = \frac{1}{m+1}$$ for all $0\leq k \leq m+1$. In particular, we have $$p_{0}p_{m} = p_{1}p_{m-1} = p_{2}p_{m-2} = \cdots = p_{m}p_{0}.$$ This gives $$\frac{p_{1}}{p_{0}} = \frac{p_{m}}{p_{m-1}}$$ for all $m\geq 1$, so that $... 2$P(Y<1|X<1)=\frac {P(X<1,Y<1)} {P(X<1)}=\frac {\int_0^{1} \int_x^{1}2e^{-x-y} dydx} {\int_0^{1} \int_x^{\infty}2e^{-x-y} dydx}$. I will let you evaluate the integrals. 2 So the log likelihood given$n$observations is $$l(\theta) = \log \prod_{i=1}^n f(y_i;\theta) \propto - \sum_{i=1}^n \log y_i - \frac{n}{2} \log \theta - \frac{1}{2\theta}\sum_{i=1}^n \log^2 y_i$$ Derive w.r.t$\theta$and set it to zero to get the first order condition: $$l'(\theta) = - \frac{n}{2\hat{\theta}} + \frac{1}{2\hat{\theta}^2} \sum_{i=1}^n \log^... 2 If m < n and f is everywhere differentiable, then measure of f(\mathbb R ^ m) is zero (see, for example, this answer - we can extend f to \mathbb R^n to apply it directly by using g(x_1, \ldots, x_n) = f(x_1, \ldots, x_m) - then \det D g is zero everywhere). So support of f(X) has zero measure and thus f(X) doesn't have density. 2 Your intuition for 100 dice is correct – the probability that at least one two appears in n die rolls is 1-\left(\frac56\right)^n, where \frac56 is the probability of not getting a two from one of the dice. 1-\left(\frac56\right)^n tends to one as n increases. For two dice the probability of at least one two works out as \frac{11}{36}>\... 2 Let's assume n die are thrown and we want the probability of at least 1 dice to show 2.$$P(\text{at least one 2 shows up}) = 1- P(\text{No 2 shows up})=1-\left(\frac{5}{6}\right)^n$$For n=100, the probability comes to be around 0.999999. So yes it's very likely when you throw a 100 die, atleast one 2 will show up. 2 The PDF tells us that X<Y a.s. so that P(Y<1\mid X=1)=0. 2 As we have 0<x<y<\infty, if we take x as dependent on y, we get -$$\iint_R f_{XY}dx dy= \int_0^\infty\int_0^yf_{XY}dxdy$$If we would take y as dependent on x, we get -$$\iint_R f_{XY}dx dy= \int_0^\infty\int_x^\infty f_{XY}dydx$$2 Yes, this is fine. To add some detail: for the second equality, you've used that \int_y p(x,y)\,dy = 1 for all x. The third is justified since f(x) is just a constant as far as \int_y cares. And for the last line, to move to the joint integral you might invoke Fubini-Tonelli. 2 We seek to find the number of solutions to x_1 + x_2 + \ldots + x_k =n in nonnegative integers. Consider n dots and k-1 dividers arranged in a row. The dividers divide the row into k sections, and the number of dots in each gives the x_i. There are n+k-1 \choose k-1 ways to arrange these, and so this is the number of ways votes can be cast. 2 You have n votes for k candidates. If you line up the votes by candidate, you'll have groups of votes for each one. (Unpopular candidates may not have any votes.) We can insert k-1 dividers in the line of votes to cordon them off by candidate. So if we have n+k-1 spaces, we'll have room for all of the votes and the dividers. Now the problem gets ... 1 Suppose X_i are iid with continuous cdf F, L_n = \min(X_1,\ldots,X_n) and U_n = \max(X_1,\ldots,X_n). Then$$\mathbb P(a \le L_n \le U_n \le b) = \mathbb P( X_1,\ldots,X_n \in [a,b]) = (F(b) - F(a))^n$$and if F corresponds to a pdf f, the joint pdf of (L_n, U_n) is$$f_{L_n, U_n}(x,y) = - \dfrac{\partial^2}{\partial x \partial y} (F(y) - F(x)... 1$EX^{-2}=100\int_0^{\infty} \frac 1 x e^{-10x}\, dx$. Since$e^{-10x} \to 1$as$x \to 0$and$\int_0^{1} \frac 1 x \, dx=\infty$it follows that$EX^{-2}=\infty$. 1$f_X(x)=\int_x^{\infty} 2e^{-x-y} dy=2e^{-x}e^{-x}=2e^{-2x}$for$0<x<\infty$. Also$F_X(x)=\int_0^{x} 2e^{-2t} dt =-e^{-2t}|_0^{x}=1-e^{-2x}, 0<x<\infty$. 1 For any$\omega$either$ \omega \in A$or$ \omega \in A^{c}$. In the first case both sides of the equation are$g(x)$and in the second case both sides are$g(y)$so the equation is true.$g(\alpha X+\beta Y)=g(\alpha)X+g(\beta)Y$is not true in general. 1 Hint:$X$also has normal distribution. Since$\beta^{t}=e^{ta}$where$a =\log\, \betaX$has same distribution as$Y+a$where$Y$has MGF$e^{t^{2}}$. Now you can use normal density function to evaluate the probability. 1 Using, as in the linked post,$p=\frac \lambda n$$$A=p^k \binom{n}{k}p^k (1-p)^{n-k}=\binom{n}{k} \left(\frac{\lambda }{n}\right)^k \left(1-\frac{\lambda}{n}\right)^{n-k}$$ Taking logarithms and expanding as a Taylor series for large values of$n$to get $$\log(A)=\left(k \log (\lambda )+\log \left(\frac{e^{-\lambda }}{k!}\right)\right)+\frac{-k^2-\lambda ^... 1 The conditional density you have found is for 0<x<y. E(Y|X)=\int_x^{\infty} ye^{x-y} dy. Integrate by parts to find the exact value. 1 There are 11 \choose 7 options for which 7 attempts have been successful, each with \left( \frac{1}{4} \right) ^ 7 \left( \frac{3}{4} \right)^4 change of occurring. So the probability of A occurring 7 out of 11 times is {11 \choose 7} \left( \frac{1}{4} \right) ^ 7 \left( \frac{3}{4} \right)^4=0.637\%. In general, you use the binomial ... 1 If you can convert the percentages of the population to integers (as in your example): Say there are c=N\cdot\frac{x}{100} individuals in the population with the characteristic; and the sample size is k=N\cdot \frac{y}{100}. Then the probability of at least one person having the characteristic would be$$1-\frac{\binom{N-c}{k}}{\binom{N}{k}}$$The ... 1 This is conditional expected probability. Given that you scratched a bunch of tickets, you know that the probability of the last one being a winner is increased only because you know that the tickets are not independently random; there is a set$1000$out of a million winners. However, if each ticket was independently random (i.e each ticket had a$0.1\%\$ ...

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