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Path A-B-C is free when A-B and B-C is free: $p_{ABC-free} = (1-p)*(1-p)= (1-p)^2$ So it is blocked: $p_{ABC-blocked} = 1-p_{ABC-free} = 1-(1-p)^2= 1-1+2p-p^2 = p(2-p)$ C is not accessible from A if ABC and AC is blocked: $p_{blocked} = p_{ABC-blocked} * p_{AC-blocked} = p(2-^2) p = p^2(2-p)$ So it is possible to move from A to C with probability ...


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In question(b), we have (150,000-2000*1)/100=130 persons who test positive after test A.So they go to Test A. In 2000 persons, persons who are actually negative but test positive are 2000*0.98*0.05=98. So when we deduct the number of false positive 130-98=32, it gives us the persons who actually has the condition.


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Note $$\frac{6!}{3!2!1!} = 120 \therefore P = \frac{1}{120}$$ For second, since each trial is independent, after $N$ trials, it's binomial distribution, and $$P(X=n)=\binom{N}{n}P^n(1-P)^{N-n}$$ For the third question, after $N$ trials, the monkey receives at least 2 banana has probability, $$Q=1-(1-P)^N-\binom{N}{1}P(1-P)^{N-1}$$ The question then asks ...


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The numerator should be $$\binom{4}{x}\binom{4}{y}\binom{4}{z}\binom{40}{6-x-y-z},$$ for any $x,y,z$ with $x+y+z\le 6$, with the obvious restrictions on $x,y,z$.


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Thanks, as for your hint, the number of marbles having labels that are at most x is x right? So P(X<=x)= (k choose x)/ (n choose k) ?


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Use the hypergeometric distribution: $$\frac{\binom{R}{x} \binom{N-R}{n-x}}{\binom{N}{n}}.$$ This yields: $$\mathbb P(\mbox{"obtaining exactly 3 black balls"})=\frac{\binom{9}{3} \binom{10}{1}}{\binom{19}{4}}.$$ $$\mathbb P(\mbox{"obtaining exactly 4 black balls"})=\frac{\binom{9}{4} \binom{10}{0}}{\binom{19}{4}}.$$ Now the probability of getting at least ...


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Unless you know something about Frankie's bias in picking numbers, you have to assume a discrete uniform distribution, which has a probability mass function of $p_Y(h)=1/(b-a+1)$ on the support $h\in\{a..b\}$. So the expectation is: $$\begin{align} \mathsf E[(Y-k)^2] & = \sum_{h=a}^b (h-k)^2/(b-a+1) & : & h\in\{a..b\}, k\in\{a..b\} ...


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$\frac{\binom{9}{3}\binom{10}{1}+\binom{9}{4}}{\binom{19}{4}}$


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Let $E(h,n)$ be the expected winnings (under the optimal stopping strategy) after flipping the coin $n$ times and obtaining $h$ heads. If you flip again, your expected earnings will be $\dfrac{1}{2}E(h,n+1)+\dfrac{1}{2}E(h+1,n+1)$. If you don't flip again, your expected earnings will be $\dfrac{h}{n}$. Therefore, you flip again iff ...


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\begin{align} (X,Y) & = \begin{cases} (a,a) & \text{with probability }p^2, \\ (a,b) & \text{with probability }pq, \\ (b,a) & \text{with probability }qp, \\ (b,b) & \text{with probability }q^2. \end{cases} \\[10pt] \text{Therefore }X-Y & =\begin{cases} 0 & \text{with probability }p^2, \\ a-b & \text{with probability }pq, \\ b-a ...


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Since there only two players, one of whom gets a point each turn, you can represent the entire game from the perspective of only one of the players. Lets do player 1: Player 1 wins if he gets two wins before five losses, and he has a probability of winning of 1/3. Therefore, the longest game will last six turns - whoever met their quota in these six turns ...


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For a trivial estimate of the coin flip game, you can't win more than $1$ (all heads), so that is an upper bound. Less trivially, we need to define our strategy. Suppose you have flipped $h$ heads and $t$ tails. Let $V(h,t)$ be the value of the game at this point. Clearly $V(h,t) \ge \frac h{h+t}$ because we can stop now and get that. If we flip ...


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Use the hypergeometric distribution: $Hg(5,R=\mbox{number of questions you learnt},12)$. The hypergeometric distribution is useful when we have a population of $N$ individuals, with $R$ individuals having the characteristic we are looking for and with $n$ as the number of individuals of a sample. For a random variable $X$ with hypergeometric distribution ...


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Without loss of generality, assume that $X$ and $Y$ are Bernoulli random variables with the same parameter $p = P\{X=1\} = P\{Y=1\}$. Thus, $X-Y$ takes on values $-1, 0, 1$. Then, we have that $$\begin{align} P\{X=1\} = p &= P\{X=1, Y=1\} + P\{X=1, Y=0\}\tag{1}\\ P\{Y=1\} = p &= P\{X=1, Y=1\} + P\{X=0, Y=1\}\tag{2}. \end{align}$$ From $(1)$ and ...


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Probability Space: Drawing 10 cards with replacement from a standard deck. Favoured Event: Drawing at least 2 queens. Complementary Event: Drawing less than 2 queens Let $Q$ be the count of queens drawn. Assuming the deck is well shuffled after each replacement, then the probability of getting a queen on any draw is independent of every other draw, and ...


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In the interests of engaging constructively with this question, here is one possible way a test could plausibly be bell-shaped: Lets say there are $N$ students in the class. We can represent their level of preparedness for the exam by a "correct answer" probability, $p_i$, which we can interpret as the fraction of questions they would get correct if they ...


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Your working is correct. The reason the number is low despite the selectivity and sensitivity of the alarm is that because there are so very few actual fires then there are still many more false alarms than true alarms.


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Looks right to me though I didn't actually check all the calculations. The low result is initially surprising, but is not uncommon in this kind of problem. It shows that even if $P(A\,|\,B)$ is very high, this does not guarantee that $P(B\,|\,A)$ is high. It is often put in the context of a medical diagnostic test for a very rare disease. The point is ...


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We draw without replacement $k$ marbles out of an urn of $n$ uniquely labelled marbles, marked from $1$ to $n$, and record the highest label of all the marbles draw as our random variable $X$. Note: The support of $X$ must therefore be: $\{k, , , n\}$ The Cumulative Probability Function of $X$ is $\mathsf P(X\leqslant x)$, and this is the probability of ...


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The tree approach is perfectly fine. Writing it with fancy symbols doesn't make it more rigorous, as it just restates what is in the tree. In either case, you will need to enumerate the possibilities and get each probability. There are exactly four ways to pass the test: Get the first three tests correct Fail one of the first three test and pass the last ...


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Apparently there are two types of standard deviation: sample standard deviation : and population standard deviation : which is confusing because it was never introduced in textbooks...


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$$ \Pr(F(X)\le y) = \Pr(X\le F^{-1}(y)) = F(F^{-1}(y))=y. $$ That works if $F$ is invertible, since invertible CDFs are strictly increasing.


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Let $X_{k}$ denote the number of boxes that are to be bought to come in possession of $k+1$ tokens, counting from the moment that one is in possession of exactly $k$ different tokens. Then $X=1+X_{1}+\cdots+X_{n-1}$ boxes must be bought. Here $X_{k}$ has geometric distribution with parameter $p_{k}=1-\frac{k}{n}$ so that ...


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Note $$\operatorname{Var}[X] = \operatorname{E}[X^2] - \operatorname{E}[X]^2 = \operatorname{E}[X^2] = \Pr[X^2 = 1] = \frac{2}{3}.$$ The first equality comes from the calculation $$\begin{align*} \operatorname{Var}[X] &= \operatorname{E}[(X-\operatorname{E}[X])^2] \\ &= \operatorname{E}[X^2 - 2\operatorname{E}[X]X + \operatorname{E}[X]^2] \\ &= ...


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It looks correct and I get the same results as your calculation with numpy. import numpy as np x = [-1,0,1] np.std(x) 0.81649658092772603 So it is probably the way you calculate it.


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Picking $m+j$ marbles from a total of $n+k$ marbles can be done on $\binom{n+k}{m+j}$ ways. However, if it is done under the restriction that $m$ are of type $A$ and $j$ of type $B$ then it can be done on $\binom{n}{m}\binom{k}{j}$ ways. So the probability of that event is: $$\binom{n}{m}\binom{k}{j}\binom{n+k}{m+j}^{-1}$$


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In a simpler case, if you want to choose k items out of n items, the number of ways to do it is $\binom nk$, now if the $k$ is equal to $n$, i.e. when you want to take all the items, there is only one way to do it, you take it all, and that'd be $\binom nn =1$. Similarly, here since you want all 4 red ones, you only have one way to do it, and then you want ...


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Hint: Let's say that $E_{i}$ denotes the event that he passes the $i$-th test. Then his chance of qualifying is: $$P\left(E_{1}\cap E_{2}\cap E_{3}\right)+P\left(E_{1}^{c}\cap E_{2}\cap E_{3}\cap E_{4}\right)+P\left(E_{1}\cap E_{2}^{c}\cap E_{3}\cap E_{4}\right)+P\left(E_{1}\cap E_{2}\cap E_{3}^{c}\cap E_{4}\right)$$


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For 1) Yes, they are both correct. 2) no, the rate is irrelevant since its a sequence of constants that disappear in the limit. Is there a counterexample that gave you concern?


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If $F$ is the CDF of a random variable then $F(x)$ stands for the probability that $X$ will not exceed $x$, i.e. $F(x)=P(X\leq x)$. If $X$ is distributed over interval $[a,b]$ then it will only take values in $[a,b]$ so that for every $x\geq b$ it is true that $X$ will not exceed $x$. Equivalently you can say that for every $x$ with $x\geq b$ the probability ...


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Look at the definition of CDF. By definition $F(x)=P(X\leq x)$. In this case, the support of the random variable is $[0,2\pi]$. So, $F\left(-\frac{\pi}{6}\right)=P\left(X\leq-\frac{\pi}{6}\right)=0$, since $X$ never attains values less than $-\frac{\pi}{6}$. Furthermore, consider a point to the right of the support of $X$, such as $F(3\pi)=P(X\leq 3 \pi)=1$, ...


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Using Markov's inequality: $$ P( |A_n-0|> \epsilon) \leq \frac{E(|A_n|)}{\epsilon}=\frac{|A_n|}{\epsilon} $$ Letting $n\rightarrow \infty$ on both sides proves convergence in probability.


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Yes, you can. And no, there is no difference (the standard mesure of the quality of the estimator is his variance, which does not change when you add something deterministic). The change between both estimators is than both of them can't be unbiaised.


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You wrote normally distributed. $23$ inches is $\dfrac{23-22.8}{1.1}$ standard deviations above the mean. What proportion of a normally distributed population is less than that many standard deviations above the mean? That's something you usually find using a table or software. (In this cases it's the proportion that's less than $23$ inches minus the ...


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$P(D) = .02$ $P(R/D) = 1$ $P(R/D') = .05$ $P(D') = 1-.02 = 0.98$ $P(R) = P(D)P(R/D) + P(D')P(R/D') = .02(1) + .98(.05) = .069$ Cost of inexpensive + follow up test, if tested positive $=10 + 0.069(100)$$ Expected cost $= 10+ 0.069*100 = 16.9 $$


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The approximation by $\Phi$ will give results that are not too unreasonable even with numbers as small as $4$ and $6$ if, but only if, a continuity correction is used. That means "$\ge 4$" is the same as "$>3$", so you say "$>3.5$" instead. That gives surprisingly good approximations even with samples as small as this, whereas textbooks usually give ...


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Defenition for probability limit: $X_n\to X$ in probability iff $\forall \epsilon>0$ $\lim\limits_{n\to\infty}P\{|X_n-X|>\epsilon\} = 0$. We should concern sequence of random numbers $X_n$ such that $P\{X_n=A_n\}=1$. As $\lim\limits_{n\to\infty}A_n=0$ then for any $\epsilon>0$ there is $n_\epsilon$ such that $|A_n|<\epsilon$ for any ...


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Let $X$ be the random variable indicating # of the toss in which the first tail occurred. Let $Y$ be a random variable which indicates the amount of money you earned. We are given that $X$ has a probability function $P_X(x) =\left( \frac{1}{2}\right)^x=\mathbb P(X=x)$. It is clear that $Y$ takes the values $2,4,8,16,32,-256$. By definition: $$\mathbb ...


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The only way to get a binomial distribution is to assume that the sampling is done with replacement. If the sampling is without replacement, the distribution of the number of defectives is hypergeometric. For your second question, if you pick $5$ items and number them, the probability the first is defective is $\frac{10}{25}$. The probability the second is ...


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The use of binomial and hypergeometric hinges on the concept of whether the samples are replaced or not. If it is replaced, then the probability of picking the second of n itmes remains the same as the first through n times that are picked while without replacement, the probability keeps reducing everytime you pick an item, in other words, the denominator ...


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Just for completeness and based on Arthur's and your comments, if the total number of possible numbers is $N=60$, the number of numbers you pick is $M=10$, the total number of lucky numbers is $n=20$ and the number of lucky numbers you pick is $k$ then using the hypergeometric distribution the probability of $k$ is $$\frac{\displaystyle {K \choose k}{N-M ...


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\begin{align} & \left(\sum_{x=1}^5 2^x\cdot\Pr(\text{winnings}= x)\right) - 256\cdot\Pr(\text{winnings}=256) \\[8pt] = {} & \left(\sum_{x=1}^5 2^x\cdot(0.5)^x\right) - 256\cdot \Pr\Big(\text{heads on all five of the first five tosses}\Big) \end{align} Now simplify the expression $2^x (0.5)^x$ and figure out the probability of getting heads on all ...


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Is the total number of cells relevant at all? Assuming you choose the cells independently, the probability of having zero $1$s in the $25$ cells chosen is $P=(0.96)^{25}$, so the probability of having at least one $1$s is $1-P$.


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The wording for the third part of the question is somewhat unclear. One way to interpret it is to ask what is the expected number of days that Mr. Li will use the machine, if he uses it twice a day every day, until he observes the third dispensing failure (and therefore lose 6 yuan, which is the first instance he loses more than 5 yuan). Under this ...


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A person can give 3 trial exams for the clearing test. In 1st attempt probability of passing is 40 , Those who have failed have 60 probability of passing the exam in 2nd attempt. Those who have failed in 2nd attempt have 20 probability of passing the exam


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Just a hint: Let $X\sim B(n,1/2)$. Define $Y=-X$. Write the cdf of $Y$. At this point the normal approximation can be used. Apply the method described here for maximum of $k$ random trials of $Y$ to get the corresponding cdf. Differentiate the cdf to get pdf for maximum of $k$ trials of $Y$. Transform this pdf to pdf of minimum of $k$ trials of $X$. ...


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So the convolution integral is: $$\int_{-1}^{1}(1-|x|)(1-|T-x|)dx$$ Now you can simply calculate this integral: $$\int_{-1}^{1}(1-|x|)(1-|T-x|)dx=\int_{-1}^{1}1dx+\int_{-1}^{1}|x|dx+\int_{-1}^{1}|T-x|dx+\int_{-1}^{1}|Tx-x^2|1dx$$ First two are easy: $\int_{-1}^{1}1dx=2$, $\int_{-1}^{1}|x|dx=1$. Third (you can eliminate $|\cdot|$ dividing integral into ...


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To determine the cdf $F_T(t)=P(T\le t)$, I would proceed geometrically. Points $(X,Y)$ come from a square with corners at $(1,0), (0,1), (-1,0), (0,-1)$. To find $F_T(t)$, draw the line $X+Y=t$ and note it is parallel to two of the sides of the square. $F_T(t)$ is the area inside the square to the left of the line. Since the line is parallel to the side ...


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$$f_Y(t-x)=1-|t-x|$$ and the integral bound should be from -1 to 1 since it is range of Sx. Then you could approach this problem by splitting (-1,1) and do the integration.



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