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Imagine that each candy has an ID number. There are $\binom{48}{5}$ equally likely ways to draw $5$ candies. Now we find the probability that Jimmy is unhappy with the bag, that is, that in his sampling he draws $0$ or $1$ strawberry. There are $\binom{6}{0}\binom{42}{5}$ ways to draw $0$ strawberry and hence $5$ non-strawberry/ There are ...

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Warm-up ($n=4$ case) Start with the case of $n=4$. We want to know $\text{Pr}\left(\sum_{i=1}^4 x_i y_i = 0\right)$. There are three cases where this sum could be zero: None of the $\{y_i\}$ equals 1, which happens with probability $p^4$. Exactly two of the $\{y_i\}$ equal 1, which follows a Binomial distribution, ie. ...

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You have the formula correct, but are misreading the table. Notice that the numbers in each row adds to 100%, but the numbers in the columns do not. Also, notice the percentages in the labels for Not Employees and Employees, respectively. What you need to do is create a table that expresses the probability of each combination. John Bush 0.208 0.442 ...

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Answer: Assume that Beavis starts the game. P(Beavis picking the red ball in his first try) = $$7/13$$ P(Beavis does not pick the red ball and Butthead picks the ball) = $$(6/13).(7/12) =$$ =$$\frac{{7\choose2}}{{13\choose2}}$$ P(Both of them do not pick the red ball and Beavis picks it at the 2nd try) = $$(6/13)(5/12)(7/11) = ... 0 What you are given is this: E = employee, N = non-employee, and P(J/N) = .32, P(B/N) = .68, P(J/E) = .61, P(B/E) = .39, P(E) = .35, and P(N) = .65. So you are to calculate: P(E/J). We use Bayes' rule to find the answer. Namely: P(E/J)*P(J) = P(J/E)*P(E) ===> P(E/J) = P(J/E)*P(E)/P(J) = = P(J/E)*P(E)/(P(J/E)*P(E) + P(J/N)P(N)) = (.61)(.35)/(.61*.35 + ... 0 Outline: Suppose that a program has n lines. Then the number X of errors in the program has Poisson distribution with parameter \lambda=n(0.015). We want \Pr(X\le 1)\le 0.10. So we want e^{-\lambda}(1+\lambda)\le 0.10. This inequality does not have a nice algebraic solution for \lambda. However, a little calculator experimentation will yield a ... 0 Let p(n) be the probability that the first player to play, as n white balls remain, wins. We have$$ p(n) = \frac{7}{n+7} + \frac{n}{n+7}(1-p(n-1))\\p(0) = 1 $$(according to the result of the first turn). From this it should be easy. 1 It sounds Like German Tank Problem: http://en.wikipedia.org/wiki/German_tank_problem: In the statistical theory of estimation, the problem of estimating the maximum of a discrete uniform distribution from sampling without replacement is known in English as the German tank problem, due to its application in World War II to the estimation of the number of ... 0 Outline: Define random variable X_i by X_i=1 if digit i is missing, and X_i=0 otherwise. Then the number N of digits missing is given by$$N=X_1+X_2+\cdots +X_9.$$By the linearity of expectation, we have$$E(X_1+\cdots +X_9)=E(X_1)+\cdots +E(X_9).$$For each i, the probability i is missing is \left(\frac{8}{9}\right)^{10}. Thus the ... 0 Answer: The total number of ways you can pair them is ofcourse 9. A can choose to dance with b or c. Suppose he chooses to dance with b. B will have a choice between a and c. She has to select a because C cannot dance with c. Similarly, If A chooses to dance with c, C will have a choice to dance with b, because B cannot dance with b. Thus either way, ... 0 The Central Limit Theorem examines sums of random variables from which we subtract the mean of the sum and then divide the whole by the standard deviation of the sum: let Y_1,...,Y_n be random variables and define S_n \equiv \sum_{i=1}^nY_i. Then the CLT examines the random variable$$Z_n = \frac {S_n - E[S_n]}{\sqrt {\text {Var}(S_n)}}$$and what is ... 1 Suppose you want the expected value of X to be some particular number \mu. For every value that the random variable Y could take, pick some probability distribution for which the expected value exists. If W is a random variable with that distribution, then let W-\mathbb EW + \mu be the value of X when Y assumes the particular value concerned. ... 1 The probability that the other team wins all 3 remaining games is 1- the probability you are looking for. 1 In general, the conditional expectation is a function of the conditioning variable E(X\mid Y) =g(Y) When this function is a constant, then (iff) E(X\mid Y) =E(X). You can't say much more, in general. Here I give an example. 0 4 (1/10)^5 (9/10)^3 + 9 (1/10)^6 (9/10)^2 + 2 (1/10)^7 (9/10) + (1/10)^8 = = 36.64 / 1,000,000 1 Nothing probabilistic here... Try X_n=1 with full probability if 4^k\leqslant n\lt2\cdot4^k for some integer k, and X_n=0 with full probability otherwise. 1 Let p(x, y) = A(x + y)^2. We know that since p(x, y) is a probability density function, we must have \int_0^1\int_0^1p(x, y) \mathrm{d}x \mathrm{d}y = 1. You need to solve this integral for A. p(y | x) := \frac{p(x, y)}{p(x)}. To calculate this, you need to find the marginal distribution of x, which you get by integrating y out of the ... 0 With the help of Did's comment "from the characterization of the conditional expectation"$$ E[Y E(X|Y)] = E[E(XY|Y)] = E(XY) $$So the second implies the first. A deleted comment said that the tower property of conditional expectation can help, but I don't know how. How about the reverse direction? 0 \frac{1\times 1\times 1\times 1\times 10\times 10\times 10\times 1\times 1\times 1\times 1\times 1\times 10\times 10\times 10\times 10}{1\times 1\times 1\times 1\times 10\times 10\times 10\times 10\times 10\times 10\times 10\times 10\times 10\times 10\times 10\times 10}=\frac{10^7}{10^{12}}=\frac{1}{10^5} 1 in 100,000 for this specific location, and ... 2 Since cov(X,Y)=E(XY)-E(X)E(Y), then for E(XY)=E(X)E(Y)to be true, X,Y are uncorrelated. 0 Hints: You have \Pr(Y=1)=0.4,\; \Pr(Y=2)=0.3,\; \Pr(Y=3)=0.2,\;\Pr(Y=4)=0.1. What are \Pr(Y\le 1),\; \Pr(Y\le 2), \;\Pr(Y\le 3),\; \Pr(Y\le 4)? What are \Pr(Y\le 0),\; \Pr(Y\le 1.5),\; \Pr(Y\le 3.7),\; \Pr(Y\le 5)? What is \Pr(Y\le y), covering all values of y? 0 Another way to solve would be The number of ways you can select a king, queen or a jack for each of the three cards=$${4\choose1}$$The total number of ways you can select three cards from the deck =$${52\choose3}$$Required Probability =$$\frac{{4\choose1}{4\choose1}{4\choose1}}{{52\choose3}} = \frac{64}{22100} = \frac{16}{5525}$$2 probability of getting a king on your first draw is 4/52. probability of getting a queen after a king on your second draw 4/51 jack after a king and a queen 4/50 So the probability of getting a king, then a jack, and then a queen is (4*4*4)/(50*51*52). There are six orders that give you a king, a queen and a jack. KQJ, KJQ, QKJ, QJK, JKQ, JQK So, we ... 1 Indeed$$P(Y>2)= 1 - [P(Y=0)+P(Y=1)+P(Y=2)]$$Then if you mean$$1-\left[\frac{(5\cdot2)^2}{2!}e^{-2\cdot5} + \frac{(5\cdot2)^1}{1!}e^{-2\cdot5} + \frac{(5\cdot2)^0}{0!}e^{-2\cdot5}\right]$$your answer is correct (be careful with the parentheses in your expression). 0 Hint: The given hint leads directly to \Pr(X_i\le y)=\frac{y-100}{100} for all y in the interval (100,200), and all i from 1 to n. By independence, it follows that$$F_Y(y)=\left(\frac{y-100}{100}\right)^n$$in that interval. For completeness, we have F_Y(y)=0 for y\le 100, and F_Y(y)=1 for y\ge 100. Then follow the given hint to find ... 1 The lack of memory of the exponential distribution can be used to produce conceptual proofs that, for every n\geqslant2, G is distributed as the maximum of (n-1) i.i.d. random variables each exponentially distributed with parameter a. Since, however, the OP failed to explain their background, here is a direct, hands-on, approach. Consider ... 0 The average arrival of customers per 8 hours is$$λ=8\cdot7=56$$So the number of customer arrivals in 8 hours periods is a Poisson process Y with an average λ=56. Therefore, the correct answer is$$P(Y=45)=e^{-56}\frac{56^{45}}{45!}$$So, assuming that the parenthesis in the numerator is a typo (see comments), your answer is correct. 0 I do not see what the Kronecker product is doing there, it is just creating a block-diagonal matrix were each diagonal block is a copy of the second factor. The component of diag(ZZ^T\mathbf1)-ZZ^T at position (i,j) is$$ \left(\sum_kZ_k\right) Z_i\delta_{ij}-Z_iZ_j $$It should now be trivial to compute the expectation and come up with zero. 1 Hint: In order for you answer to be correct you also need to add the following term in the numerator$$\begin{align*}&\phantom{\,\=}P(m \text{ events in $t$ hours | $n$ events in $T$ hours})= \\ \\& =\frac{P(m \text{ events in $t$ hours, $n-m$ events in $T-t$ hours})}{P(n \text{ events in $T$ hours})}=\\\\& =\frac{P(m \text{ events in $t$ ...

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Hint: Let $B$ be the event "$n$ events in the first $T$" and $A$ the event $m$ events in the first $t$." We want $\Pr(A|B)$, which is $\frac{\Pr(A\cap B)}{\Pr(B)}$. To compute $\Pr(A\cap B)$, note that for this event to occur we need $m$ events in the first $t$ hours and $n-m$ additional events in the remaining $T-t$ hours. Thus you need to multiply your ...

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Poisson Distribution is given by $$f(y) = \frac{\lambda^ye^{-\lambda}}{y!}$$ For part A the Question asks for the probability that NO MORE THAN THREE arrive. So that is simply $$P(0 ~~arrive) + P(1 ~~arrives) + P(2 ~~arrive) + P(3 ~~arrive)$$ What you've solved for instead is $$P(more ~~than ~~3 ~~customers ~~arrive)$$ Get rid of the 1 minus and you're good ...

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The reason that 'all random numbers have to come from a distribution' is the same as the reason that 'all maps from $\mathbb{R}\to\mathbb{R}$ have to come fron a function'; namely, that one defines the term itself (be it function or random — more properly probability — distribution) as any entity that satisfies all of the hypotheses one would ...

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Add up the first 14 numbers. Then, whatever the remainder of the result modulo $10$, adding the 15th number will give each of the possible remainders with equal probability. So the answer is $0.1$.

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We often think of random numbers "coming from", or more aptly, "belonging to" a distribution, because this is useful, and this is how we generate them computationally. However, it is perhaps more useful to think of random numbers as an endemic property of the universe, and the distribution is what we use to describe them. For example, suppose you wrote ...

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Try to read it like $$\sum_{i_1}z^{i_1}...\sum_{i_n}z^{i_n}=\sum_k(\sum_{i_1+...+i_n=k}1)z^k$$ If you going to write out the LHS then you want the know the coefficient in front of $z^k$ you want to know how manny different ways $z^{i_1}\cdot...\cdot z^{i_n}=z^{i_1+...+i_n}=z^k$ so how many ways $i_1+...+i_n=k$. Which is what we have on the RHS. Addition: ...

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I'm less skeptical than Andre about the normal approximation in this case-- I think it might work out well. Let's see. For the normal approximation with correction for continuity, we want $P(X > 45.5)$, where $X$ has a Normal distribution with mean 35 and standard deviation 5.400617. (Andre is correct, your original calculation of the standard deviation ...

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The number of unacceptable chocolate bars in a sample of $1000$ is a random variable $X$ which has the binomial distribution with parameters $n=1000$ and $p=0.10$, in symbols $$X \sim \mathrm{Bin}(n=1000,\,p=0.10)$$ To calculate the required probabilities we will use the normal approximation to the binomial distribution, that is $$X \sim ... 0 The important fact is that in your population, all remainders on division by 10 are equally probable-there are 150 with each ones digit. If you choose the first two and add them, you define what the ones digit of the last needs to be. It will be that with probability \frac 1{10}. This is really a restating of frabala's answer without the language of ... 2 Say x_i, i=1,\ldots,15 are 15 3-digit numbers. The probability that 10 divides their sum is equal to the probability that$$\sum_{i=1}^{15} x_i=0 \pmod{10}$$But this corresponds to choosing 15 1-digit numbers (zero's and repeats allowed). For each x_i the probability that x_i=z_j \pmod{10} (for some z_j\in\{0,\ldots,9\}) is \frac{1}{10}. Then you ... -1 Consider only last digit (frabala explains it and it is pretty obvious too): a_1+a_2...a_{15}=0,10,20...130 1 solution for 0 Note that 0\leqslant a_i \leqslant 9 Number of solutions for e.g. 10 Coefficient of x^{10} in (1+x^1+x^2...x^9)^{15} Explanation : What happens when while opening that messy ^15 you multiply x^somethings and get ... 0 Well, the smallest sum is 100+101+102+...+114 = 100\cdot 15+\frac{14\cdot 15}{2} = 1605, while the greatest sum is 999+998+997+..+985 = 1000\cdot 15-\frac{15\cdot 16}{2} = 14880, so all the possible sums are 14880-1605+1 = 13276, while the sums divisible by ten are ... 0 The answer I get is \dfrac{1}{15}. I used a Venn diagram but here's a mathematical proof. First, Pr(h)=0.6 so Pr(h^c)=0.4. Also, Pr(p|h^c)=0.7 so P(p \cap h^c)=Pr(p|h^c)\cdot Pr(h^c)=(0.7)\cdot(0.4)=0.28. Further, Pr(p)=Pr(p \cap h)+Pr(p \cap h^c). So Pr(p \cap h)=0.3-0.28=0.02. Finally, Pr(h|p)=\dfrac{Pr(h \cap ... 0 The singletons look to me like an unnecessary detour. If F were generated by countably many of its elements, then it would also be generated by a family G of countably many countable sets, because you could just replace any co-countable sets among the original generators by their complements. The union U of all the generators in G is a countable ... 3 Suggestion: Forget about the pigeons and holes and think about balls and bins instead! (an equivalent but more common model in the probabilistic literature). The following is an outline of a proof you may find here: Let C(n,m) be the collision probability when randomly throwing m < n balls into n bins (or m pigeons into n holes?!). Then we ... 0 You can divide people in 4 categories: HP happy and poor HR happy and not poor SP not happy and poor SR not happy and not poor you know HP+HR, SP, and HP+SP. You are asked to find HP. 0 A good start is to write down the information you have in probability notation:$$P(H)=0.6P(P|H') = 0.7P(P) = 0.3$$You want to find:$$P(H|P)= \frac{P(H \bigcap P)}{P(P)} $$Since,$$P(P|H')= = \frac{P(P \bigcap H')}{P(H')} =\frac{P(P \bigcap (1-P(H)))}{1-P(H)} $$-1 There is a Markova property as \min A_i\sim exp And it is correct with your step as\max A \le min\implies\max=min. I think the difficulty is how to get the distribution function from the wrok you have done 1 I have satisfied myself by an argument like the following. May be it isn't quite what you need, but it does avoid the use of Stirling's formula, which makes it easier for people like me to follow. Assume you have placed \sqrt n pigeons, and no collisions took place. Then a fraction of 1/\sqrt n of the holes are occupied. This means that the probability ... 2 Hint: There are 2^4 equally likely outcomes. Finding the number of "favourable" outcomes is good. There is also a perhaps familiar formula ("binomial distribution") This problem is exactly the same as: "We toss a fair coin 4 times. What is the probability of exactly 3 heads or exactly 3 tails?" 1 A tedious but systematic way forward would be to use the rule P(X)=P(X\cap Y)+P(X\cap Y^c) repeatedly until every term on both sides of the equation is the probability of an intersection of either X or X^c for each free set X in the identity. In your example, P(A) becomes$$P(A\cap B\cap C)+P(A\cap B^c\cap C)+P(A\cap B\cap C^c)+P(A\cap B^c\cap ...

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