New answers tagged

1

You are correct, in the solution we require $x>a$. It is legitimate to replace $P(X>a\mid X=x)$ with $1$ only if $x>a$; otherwise $P(X>a\mid X=x)$ equals zero, and the resulting conditional density is zero as well. As a check, the conditional density integrates to 1 provided it is zero when $x\le a$.


3

Ad (a): You have calculated the probability that a specific order is selected. But there are $7!$ possible orders. Ad (b): You are right to use the converse probability. Just use the hint for (a). Ad (c): The result of the selected balls look like $AABCDEF$. It can be arranged in $\frac{7!}{2!}=21$ ways. $AA$ can be $one$ out of $seven$. Given $AA$, ...


2

a. Each time we pick one of the seven balls, so there are totally $7^7$ ways to do it. If each ball is selected exactly ones, it is equivalent to permutating the numbers $\{1,2,3,4,5,6,7\}$. For example, if the permutation is $\{1,3,5,4,2,7,6\}$, then the ball labelled by $'1'$ is selected at the first time, the ball labelled by $'3'$ is selected at the ...


3

(a) - you need the probability at each step of the next ball being none of hte preceeding balls. Only when you get to the last ball is that as small as $\frac 17$. $$ 1.\frac 67.\frac 57.\frac 47.\frac 37.\frac 27.\frac 17 = \frac{7!}{7^7} \approx 0.00612$$ (b) As you correctly infer, this is $1-$ the result above. $$1-0.00612 = 0.99888$$ (c) This is ...


6

There are $7!$ ways of picking $7$ different balls (in order), and $7^7$ ways of picking $7$ ones with possible repetition (in order). So the answer to the first question is $\dfrac{7!}{7^7}$. This is the complement of the first one, so the answer is $1 - \dfrac{7!}{7^7}$. There are $7$ ways of choosing the one ball that will not be included in the ...


5

What is the probability that each ball is selected exactly once? $$\frac{7!}{7^7}$$ What is the probability that at least one ball is not selected? $$1-\frac{7!}{7^7}$$ What is the probability that exactly one ball is not selected? $$\frac{\binom{7}{6}\cdot\binom{6}{1}\cdot\frac{7!}{1!\cdot1!\cdot1!\cdot1!\cdot1!\cdot2!}}{7^7}$$


1

I agree with both your calculations and get $\frac{1892}{23205}$ for both of them.


0

Scenario $B$ is equivalent to a scenario where each die has success probability $\frac23$ on the first roll and $\frac16$ on all further rolls, and is rerolled as long as it succeeds. We can use this to express the distribution as a single convolution, instead of the $(x+y)$-fold convolution that you wrote. The probability for $k$ of $m=x+y$ dice to succeed ...


1

I think you meant standard deviation is b? Then Variance=$b^2$ Because $Var(X)=E[X^2]-(E[X])^2$, and $E[X]=\mu=a$ So $E[X^2]=a^2+Var(X)=a^2+b^2$


2

This is purely a matter of definitions: recall that $Var(X)=E(X^2)-E(X)^2$. Try to conclude from there based on what you know. Also they might have meant that variance is actually $b^2$, not $b$.


2

There's a subtle flaw in your reasoning. You're effectively conditioning on which non-common gem type comes first. Your calculation for phase $2$ is correct: The non-common gem type already seen in phase $1$ has become irrelevant, and only the relative probabilities of the other two gem types matter. But this doesn't work in phase $1$, because the length of ...


3

Two possible techniques. First technique: use other people's work. The distribution in question appears to be, from @BGM's comment, a multivariate hypergeometric distribution, which is a very enjoyable name to pronounce very quickly 5 times in a row. More precisely, $(X,Y,k)\sim \operatorname{MultivHypergeom}_3(\underbrace{(\gamma N,\gamma N,(1-2\gamma) ...


0

Let $$ P(x) = (q_1 + p_1 x)(q_2 + p_2 x) \cdots (q_n + p_n x)$$ with $q_i=1-p_i$. The coefficient of the term $x^k$ comes from picking, from each factor, $k$ terms of the form $p_i x$ and $n-k$ termns of the form $q_i$ . Hence $$ a_{n,k} = \sum_{S_{n,k}} \prod_{j\in S_{n,k}} p_j \prod_{ \ell \notin S_{n,k}} q_\ell$$ where $S_{n,k}$ are all the subsets ...


0

You can use Dynamic Programming as $N$th turn's outcome is mutually independent to $N-1$th and there are two possible cases here : $K$ heads already came in $N-1$ turns $K-1$ heads already came in $N-1$ turns $dp[i][j]$ : probability of getting $j$ heads in $i$ trials. So, $dp[n][k] = dp[n - 1][k]\cdot (1 - P[n]) + dp[n - 1][k - 1]\cdot p[n]$


0

You can use Dynamic Programming as Nth turn's outcome is mutually independent to N-1th and there are two possible cases here : 1. K heads already came in N-1 turns 2. K-1 heads already came in N-1 turns dp[i][j] : probability of getting j heads in i trials. So, dp[n][k] = dp[n - 1][k]*(1 - P[n]) + dp[n - 1][k - 1]*p[n]


0

$\sum_\limits{x=0}^{99} P(x)\sum_\limits{y=x+1}^{100} P(y) = P(Y>X)$ $(\sum_\limits{x=0}^{99} P(x)\sum_\limits{y=x+1}^{100} P(y))y = E[y|y>x]P(y>x) = \frac 12 E[Y]$ I am having a hard time remembering exactly why this is....more later... $\frac {E[Y]}{2P(Y>X)} = \frac {E[Y]}{1-P(X=Y)} = \frac{50}{1-0.0563} = 52.98$ If we relax the condition ...


0

EDIT: I made the mistake of including the case where $|X_1-X_2|=0$ in $E(|X_1-X_2|)$, when that has no impact on it. I've changed that now. I have a closed form expression that is somewhat inelegant, if anyone is interested. The distribution is clear but I'll proceed as if we didn't know what it was.$$$$ The space of events is $\{0,1\}^{200}$. If $X_1$ is ...


1

I've got a strange result for expectation, so feel free to downvote it: At first, lets split $k$ balls taken into two parts - one with red and blue balls, and other - with green balls. So we get the random variable $U$: $$P(U=u) = Hypergeometric(u, k,2\gamma N,N) $$ which means that drawing $k$ balls from the box with $N$ balls that contains $2\gamma N$ ...


2

Interpretation of the question: $N$ follows Poisson distribution. $X_i$'s are IID. \begin{align*} \mathbb{E} (min(S_N,3)) &= \sum_{n=0}^{\infty} \mathbb{E}(\min(S_N,3)|N=n)Pr(N=n)\\ &=\sum_{n=0}^{\infty} [Pr(S_n=1)+2Pr(S_n=2)+3Pr(S_n \geq 3)]Pr(N=n) \end{align*} $Pr(N=n)$ can be computed using Poisson distribution. $Pr(S_n=0)=\begin{cases} 1, ...


4

To approach the problem via normal approximations, let's first think about normal distributions in general to understand the distribution of the maximum we assume that $X_1,X_2$ are independently distributed as normal variables with mean $\mu$ and standard deviation $\sigma$. We see that $$P(\max (X_1,X_2)<\mu+t\sigma)=P(X_1<\mu+t\sigma)\times ...


1

Interpretation of question : $S_N$ follows Poisson distribution. \begin{align*} \mathbb{E}(\min(S_N,3))&=Pr(S_N=1)+2Pr(S_N=2)+3Pr(S_n\geq 3) \\ &=Pr(S_N=1)+2Pr(S_N=2)+3(1-Pr(S_N=0)-Pr(S_N=1)-Pr(S_N=2)) \\ &= 3-3Pr(S_N=0)-2Pr(S_N=1)-Pr(S_N=2) \end{align*} Each of the probability term can be computed using Poisson distribution.


5

Let $X$ be the random number of heads flipped by Player $1$ and $Y$ be the random number of heads flipped by Player $2$. Suppose $X$ and $Y$ are IID binomial random variables with common parameters $n$ and $p$. Then the desired expectation is $$\operatorname{E}[\max(X,Y)] = \sum_{x=0}^n \sum_{y=0}^n \max(x,y) \binom{n}{x} p^x (1-p)^{n-x} \binom{n}{y} p^y ...


4

The other answers are great, but with a slightly different approach from your attempt (I presume you edited in your attempts after some of the answers are posted). I'd like to point out some important things in your post: "And" vs. "Or" One fundamental flaw in both approaches is that you are adding probabilities instead of multiplying them. When you are ...


2

I don't see an easy way. For your example, there are ${100 \choose 5} = 75287520$ possibilities for $B$. Of those, ${96 \choose 5}=61124064$ have no neighboring pairs. To count the number with neighboring pairs, you have $2$ choices pair at the end, then ${95 \choose 3}=138415$ ways to choose three non-neighbors out of what is rest, plus $97$ ways to ...


2

Add an $(n+1)$-th draw. Glue the ends of the interval together to form a circle, and cut the circle at the location of the $(n+1)$-th draw to form an interval again. The remaining $n$ draws are independently uniformly distributed on that interval. By symmetry, the $n+1$ segments between the $n+1$ draws all have the same expected length. Thus the expected ...


0

You can find random variables $X, Y, Z$ such that every pair of variables is independent, yet the trivariate joint does not factorize in any way. Continuous example: Divide up the unit cube into eight congruent subcubes of side length $1/2$. Select four of these cubes: Subcube #1 has one vertex at $(x,y,z)=(1,0,0)$, subcube #2 has one vertex at $(0,0,1)$, ...


1

Just for your reference as an alternative solution, though more complex. Let $X$ be the number of coin tosses until we get a success. Then $$ \Pr(X = i) = q^{i-1}p $$ The corresponding generating function is $$ f(x) = \sum_{i=1}^\infty q^{i-1}p\cdot x^i = \frac{p}{q}\cdot\sum_{i=1}^\infty (qx)^i = \frac{p}{q} \cdot \frac{qx}{1-qx}=\frac{px}{1-qx} $$ Now ...


1

$\{Y=n+i\}$ is the event that among $n+i-1$ experiments there were $i$ failures, and secondly also that the $n+i$-th experiment was a success. There are $\binom{n+i-1}{i}$ ways of selecting $i$ experiments out of $n+i-1$. Each of these selections goes along with a probability of $p^{n-1}q^i$ of occurring. The $n+i$-th experiment has probability $p$ to ...


1

Use the definition of weak convergence you have with $\chi:=f\mathbf 1\{|f|\lt R\}$ where $R$ is fixed. Then we have, by Cauchy-Schwarz inequality, $$\left(\mathbb E\left[f\cdot f\mathbf 1\left\{|f|\lt R\right\}\right]\right)^2=\lim_{T\to +\infty}\left(\mathbb E\left[f_T\cdot f\mathbf 1\left\{|f|\lt R\right\}\right]\right)^2\leqslant \liminf_{T\to ...


0

Let $A$ be the $n\times (n+1)$ matrix obtained by attaching a column of ones to $P-I$, and let $x$ be a row vector of length $n$ with $AA^T x^T={\bf 0}.$ Then $\| xA\|^2=xAA^Tx^T=0$ so that $xA={\bf 0}.$ This shows that $x$ is an invariant measure for the Markov chain. But since $P$ is irreducible, all invariant measures are of the form $\lambda \pi$, ...


0

This is how I would approach the question (Even if I made a mistake, you'll get the idea): At first, we should choose the model. Lets do hypothesis testing for binomial distribution. Our device has binomial distribution with some constant probability of success $p$ and probability of failure $1-p$. We are checking hypothesis $H_0: p \geq 0.99$ vs $H_1: p ...


0

As lulu indicated in a comment, you need to count the number of combinations of roots with repetition. You have $d$ roots to distribute over $p$ bins; the number of ways to do this is $\binom{d+p-1}d$, so the probability is $$ \frac{\binom{d+p-1}d}{p^d}\;. $$ For $d\ll p$ this is roughly $\frac1{d!}\frac1p$, so about a fraction $\frac1{d!}$ of all monic ...


0

You reach into a bag and pull two marbles. You look at the first and see that it is black. There is a 1/2 chance that the other is red. You reach into a bag and pull two marbles. Your friend tells you whether or not you have 2 red marbles. He says you don't have 2 red marbles. The chance you have 2 black marbles isn't 1/2 or 1/4, it's 1/3. The ...


0

You've confused things with clutter.   Keep it clean. We know $T_1 \sim \mathcal{Geo}_1(p)$ so $\mathsf E(T_1) = 1/p$ We know $T_2\mid T_1{=}k~\sim~\mathcal{Geo}_1(p/k)$ so $\mathsf E(T_2\mid T_1)= T_1/p$ Finally we just use the Law of Iterated Expectation to put it together: $$\begin{align}\mathsf E(T) =&~ \mathsf E\big(\mathsf E(T_1+T_2\mid ...


1

Both "Cov" and "Var" are used to represent the covariance matrix of the vector $\bf X$. See for example this Wikipedia remark: Nomenclatures differ. Some statisticians, following the probabilist William Feller, call the matrix $\Sigma$ the variance of the random vector X, because it is the natural generalization to higher dimensions of the 1-dimensional ...


1

By definition, $L$ is the max of $U$ and $1-U$. The max of two numbers is less than something if and only each of the numbers is less than that something. By this argument, we've shown: $$\{L<t\} = \{U<t, 1-U<t\}.$$


1

$U$ is the break point which lies uniformly distributed on $(0;1)$. $L$ is the length of the longer side of the break.   This is somewhere on $[\tfrac 1 2; 1)$ . When the break point $U$ is less than $1/2$ the length of the longer stick is $1-U$, and other wise it is $U$. So, for any $\tfrac 1 2\leq l\leq 1$, then the longer length being less than ...


1

The probability of selecting $y_1$ of $N_1$ white balls in a sample of $n$ from $N$ (where $3\leq n< N_1$ is: $$\mathsf P(Y_1=y_1) = \dfrac{\binom{N_1}{y_1}\binom{N-N_1}{n-y_1}}{\binom{N}{n}}$$ Which is a Hypergeometric Distribution, whose mean and variance you should know; or you can obtain by using indicators. Let $Y_{1,i}$ be the indicator that the ...


0

You have to read carefully. Here, the probability of a girl means the probability of at least one girl. You have $$P(\hbox{one child is a girl}) = 1 - P(\hbox{no child is a girl}) = 1 - P(\hbox{two boys}) = \frac{1}{4}. $$ Now you are calculating the conditional probabilty $$P(\hbox{two girls}| \hbox{at least 1 girl} = {P(\hbox{two girls})\over P(\hbox{at ...


0

I feel like people are always left with a feeling of "sure, I see how you can obtain that result but why isn't my interpretation correct?" Here's my latest attempt to address this issue. http://math.stackexchange.com/a/1801261/342573.


0

You have taken ten samples, each with an independent probability for success of $0.30$. The probability for no failures is the probability for all $10$ being successes.   What is that? The probability for at least one success is the probability for not "no failures".   (Hint: Law of Complements.)


1

I believe you're correct, the answer 1/3 is correct based on the wording of your problem. According to your problem, there is no specification on whether the boy is the oldest or youngest and therefore, the boy's siblings can either be a older sister, younger sister, or a brother. Notice that the boy can't have an older or younger brother, because the ...


0

You need to state your assumptions. However, below is a hint for a solution given typical assumptions. Hint: To compute the probability of no failures you may want to look at the Binomial random variable. The second question is the typical kind of question in which it is easier to compute the probability of the complement first and then compute ...


26

Both probablyme and carmichael561 have given a good approach to the problem, but I thought I'd point out why the solutions given by you and your classmates (?) are erroneous. The problem common to both approaches is that they neglect the probability that you die from earlier jelly beans. You take the first jelly bean, and you have a $1/10$ probability of ...


20

So you live if you do not choose a deadly jellybean :) And we die if we select at least one deadly bean, so I think it goes as follows $$P(\text{Die}) = 1-P(\text{Live}) = 1-\frac{\binom{10}{0}\binom{90}{10}}{\binom{100}{10}}=0.6695237889.$$ In this case, we literally have good beans and bad beans, and we select without replacement. Then the number of bad ...


13

It's easier to find the complementary probability; i.e. the probability that all $10$ jellybeans are safe. There are ${90\choose 10}$ ways to choose $10$ safe jellybeans, and ${100\choose 10}$ to choose any $10$ jellybeans, so the probability that you survive is $$ \frac{{90\choose 10}}{{100\choose 10}}=\frac{90\cdot 89\cdots 81}{100\cdot 99\cdots 91}$$ The ...


4

Calling the persons $P_i$ there is a probability of $7^{-5}$ that person $P_i$ steps out on floor $i$ for every $i\in\{1,2,3,4,5\}$. There are $5!$ rearrangements for the persons that result in a step out of exactly one person at the floors $1,2,3,4,5$. Moreover there are $\binom75$ ways to choose $5$ floors. So we end up with probability ...


4

This is easiest if we treat the people as being distinguishable. There are $7$ floors each person can go to, and since the floors are chosen independently, there are $7^5$ total ways for the people to get off of the elevator. There are $\binom{7}{5} = 21$ ways to choose different floors for all $5$ people, and having chosen these floors, there are $5!$ ...


3

Hint: We use inclusion-exclusion. $$P(A\cup B\cup C) = P(A)+P(B)+P(C)-P(AB)-P(AC)-P(CB)+P(ABC)$$ We are told that $AB = \varnothing$ which implies that $ABC = \varnothing$, and $$P(AB) = P(ABC) = 0.$$ So we have $$P(A\cup B\cup C) = P(A)+P(B)+P(C)-P(AC)-P(CB).$$


0

Your expression looks not quite right. If $r=n$ then that term for the probability is negative. If $n<3$ then the whole thing is nonsensical. One can see for $n=r$ and $n=r-1$ the term for the conditional probability will be zero, because there aren't enough magenta balls. So I'd pick a reasonable number ($n=6$, say) and write out the probabilities ...



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