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## Hot answers tagged probability

4

Suppose we have a discrete random variable $Y$ with $\mathbb P(Y=n) = \frac{1}{n(n+1)}$ for all positive integers $n$ then $\mathbb P(Y\ge n) = \frac1n \le \mathbb P(X \ge n)$ for all positive integers $n$ and $\mathbb P(Y\ge x) \le \mathbb P(X \ge x)$ for all real $x$ so $X$ has weak first-order stochastic dominance over $Y$ making $\mathbb E[Y]\le ... 4 If the random integer is uniformly distributed across the range$1$to$100$then the probability of guessing the integer is always$1$in$100$regardless of any bias in the guess. So, yes, a random guess will succeed in one trial in$100$on average, but so will guessing$39$every time, or any other strategy. A random lottery ticket is no more or less ... 4 Hint: Try proving that after an odd number of days there will always be exactly one person with the correct hat. 4 We have$\binom{5}{2}$possible pairs and the sum is even if we choose the unique pair$(2,4)$or two odd pairs that is$\binom{3}{2}$, therefore $$P(\text{even sum})=\frac{\binom{3}{2}+1}{\binom{5}{2}}$$ 4 We try to pick three guilty criminals in a row. Pick a criminal. What's the chance they are guilty? (It's$\frac{4}{10}$, you have already mentioned this yourself). So after you have taken this criminal out of the queue, select a second criminal. What is the chance that this second criminal is also guilty? (Hint: it's not$\frac{4}{10}$.) Multiply those ... 3 Yet another method: the table of possible birthday pairs: There are$365\times 365$cells (birthday pairs), and$365of them (diagonal ones) are pairs with birthday coincidence. So we get the ratio $$\dfrac{\color{orange}{required\;\; pairs}}{\color{gray}{pairs\;\; at \;\; all}} = \dfrac{\color{orange}{365}}{\color{gray}{365\times 365}} = \dfrac{1}{365}... 3 You're dealing with multiple issues here - one deals with an expected value calculation given truncation (probability), and another deals with estimation of parameters using data (statistics). Given that you haven't specified a method for estimation of your parameters, my guess is that there's a lot of background that you're missing. First of all, assuming ... 3 A randomly picked number is just as likely to match another randomly picked number as does a human-picked number. Thinking that a computer-picked number has a better chance of winning than does a human-picked number because of the fact that 70% of all wins are scored by a computer does not mean that computers are better at this, and commits the base rate ... 3 The probability that a given person raises his hand is the probability that someone else at the party has the same birthday as he does:$$1-\left({364\over365}\right)^{24}$$The expected number of hands raised is$$25-25\left({364\over365}\right)^{24}\approx1.593,$$by linearity of expectation. 3 Let A be the event that all three chosen numbers are at least \frac{2}{3}. Let B be the event that the first number chosen is the smallest of the three, is less than \frac{2}{3}, and is within \frac{1}{3} of the others. The probability we want is, by symmetry, P(A)+3P(B) (the extra factor of 3 from choosing which number is smallest). We have ... 3 You pick Door 1. The prize is behind that door 50% of the time. This means the prize is behind either door 2 or door 3 50% of the time, as well. So, once the announcer reveals one of the doors that does not contain the prize, the remaining door also has a 50% chance of being the correct door. So, whether the contestant sticks with door 1 or changes to the ... 3 Let P_n denote the probability of scoring on the nth draw. P_n = {n \choose 6}/ {48 \choose 6} -\sum_{i=5}^{n-1} P_{i} for n >5. With P_0=P_1= \cdots=P_5=0. I'm not familiar with the game, but I believe in order to "score" you must get all 6 digits correct. Define N_n as the number of ways to win on the nth draw it is: N_6 =1 N_7 =7-1=... 3 What you have done is correct. To see that EX_n \to \infty just note that (by direct evaluation of the integral) EX_n \geq \int_0^{1-\frac 1 n} \frac 1 {1-\omega} d\omega =\ln n \to \infty. \int_0^{1-\frac 1 n} \frac 1 {1-\omega} d\omega =-\ln (1-\omega)|_0^{1-\frac 1 n} =-\ln (\frac 1n)=ln \, n 3 After the first man takes a hat that is not his, here is what remains: n - 2 hats belonging to men still in the room. n - 2 men whose hats are still in the room. One hat (the "extra" hat) that belonged to the first man (who is not in the room). One man (the "extra" man) whose hat is no longer in the room because the first man took it. The "extra" man ... 3 P(X=Y)= \sum\limits_{k=0}^{\infty} p(1-p)^{k} e^{-\lambda} \frac {\lambda ^{k}} {k!}. This is nothing but pe^{-\lambda} \sum\limits_{k=0}^{\infty} \frac {r^{k}} {k!} where r=(1-p)\lambda. Can you take it from here? 3 The first one simply means P(X=0) given Y=2 and so it is equivalent to \dfrac{P(X=0\bigcap Y=2)}{P(Y=2)}. The second one simply requires you to consider the cases where XY=0 and add up the corresponding probabilities in the table. The third one is similar, though this time it involves an inequality. And finally, just find the values of X and Y ... 3 You have a Markov chain. Let E(n) be the expected number of rolls to get a 20 given that your are rolling an n sided die. You can set up a set of simultaneous equations. The first is$$E(20)=\frac 1{20}+\frac {18}{20}(E(20)+1)+\frac 1{20}(E(12)+1)$$because you have \frac 1{20} chance of rolling 20 and being done, \frac {18}{20} chance of ... 3 Note: @ThomasAndrews gave a correct solution in the comments of the main post. His solution involves a double summation, while mine below only involves one or two single summation(s), and also attempts to highlight the symmetry aspect. However, I don't know how to turn any of them into a closed form. First, the case when A throws N+1 coins vs B throws ... 3 You are a little off. The probability of drawing four cards, none of which is a face card, is$$\frac{40 \times 39 \times 38 \times 37}{52 \times 51 \times 50 \times 49}Another way to look at the problem, which leads to the same answer, is that there are \binom{52}{4} four-card hands, all of which we assume are equally likely. Of these, \binom{40}{4}... 3 I assume the real question is this: if you randomly line up these 10 people, what is the probability that the five girls end up in front of the 5 boys? Well, there are {10 \choose 5} ways to pick the 5 positions for the girls, and having them all in front is exactly 1 of those, so the probability is \frac{1}{{10 \choose 5}} And just to be clear:... 2 This is known as the probability integral transform. For 0<t<1 we have \begin{align} F_Y(t) &= \mathbb P(Y\leqslant t)\\ &= \mathbb P(F_X(X)\leqslant t)\\ &= \mathbb P(X\leqslant F_X^{-1}(t))\\ &= F_X(F_X^{-1}(t))\\ &= t, \end{align} so that Y is uniformly distributed over (0,1). Note that when X is not continuous the map ... 2 The variable -1/X does not have a finite expected value. Note that X could be 0 with positive probability, whence 1/X will be infinite (or undefined, whatever you prefer) with positive probability. Even if you could make sense out of the answer, it would not be -1/\lambda. Note that f(y) = -\frac 1 y is a strictly concave function, so by Jensen'... 2 The mixing loses information. A simple example: Suppose X\in\{0,1\} is a fair coin (Bernoulli with p=1/2), hence H(X)=1 bit. Let Z=X and W=1-X ("negated"). Let Y be a fair mixture (\alpha=1/2) of W and Z. It's clear that I(X;Z)=I(X;W)=1 bit: knowing either Z or W gives us total information about X. But knowing their mixture Y ... 2 Your first approach is correct. The way to get an answer to this is to look at the problem inductively. You can pair the first woman off with 10 men. The second woman could then choose from 9 men. Going on like this you would conclude that the tenth woman could choose from 1 man. Hence your answer is going to be 10 \times 9 \times \dots \times 1 = 10!.... 2 I am giving a proof for the continuous case. Integrating by parts in the definition of expected value and observing that P(X<-t)=0 for any t>0 you have \mathbb{E}[X]=\int\limits_0^{\infty}P(X>t)\,dt\ge \int\limits_0^{\infty}\frac{1}{t+1}\,dt=\infty $$(since P(x>t)\ge P(x>{\rm{ceiling\ of}}\, t) and {\rm{ceiling\ of}}\, t\le t+1 2 (Just a tip: if the problem is asking you whether a statement holds rather than asking you to prove a statement, most likely that statement doesn't hold :-) ) Hint: Let X be a a random variable that takes on the values \pm 1 with equal probability on your favorite probability space, and let Y=X^2. What is \sigma(Y)? 2 Assume your brother brings x+2 red and y blue marbles, and leaves 2 marbles, where 0\leq x+y\leq 6. Then the probability that those two marbles are both red is: {\binom{x+2}{2}/\binom{x+2+y}{2}}, which is claimed to be 1/2$$\dfrac{(x+2)(x+1)}{(x+y+2)(x+y+1)}=\dfrac{1}{2}$$Find the integer solution. Hunt and seek is a viable method. Note: ... 2 Since you have already found your answer, here are just some follow-up comments. You are right the key is to solve {x \over b} {x -1 \over b-1} = \frac12 or equivalently, b(b-1) = 2x(x-1) for positive integers b, x. This is one equation with two unknowns, so it can have multiple solutions, and indeed it does: the first few pairs are (b,x) = (4,3), (... 2 This is the solution for the 2D problem Fix A to be a distance r from the center of the circle (i.e. fix |A| = r). Note that the probability density function of |A| is$$f_{|A|}(r) = \frac{2\pi r}{\pi} = 2r \qquad r \in [0,1]$The$2\pi r$is the "area" of the circle representing all possible$A$for which$|A| = r$; and the$\pi$in the ... 2 It's$1/365$because it doesn't actually matter when the first person's birthday is. No matter when their birthday is, there are$365$equally likely days on which the second person can have their birthday, which is how we get$1/365\$.

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