New answers tagged

0

Another way of saying this: Given that the person you picked is male, the probability of them being employed is $0.8$. Given that the person you picked is female, the probability of them being employed is $0.5$. The probability of the person you pick being male is $0.5$. Same thing for female. So the probability is $$P = P(\text{male})P(\text{employed, ...


0

The answer is: P(guy|employed) = P(man).P(employed|male) + P(woman).P(employed|woman) = (0.5).(0.8) + (0.5).(0.5)


2

Let's take it from your last equation, where you assume conditional independence: $$\mathbb{P}(\text{dot-dot received} \mid \text{dot-dot sent}) = \left[\mathbb{P}(\text{dot received} \mid \text{dot-dot sent}) \right]^2.$$ To make things clearer, we will number the signals: $$\mathbb{P}(\text{dot1-dot2 received} \mid \text{dot1-dot2 sent}) = \mathbb{P}(\...


1

If you consider this in the context of the four outcomes, and given the independence of signals, it's perhaps easier to understand. Certainly independence means that the signal received can only depend on the signal sent, not on the next or previous signal. It also means - as I read it - that the signal sent also does not depend on preceding or following ...


2

The error comes right at the last step. $$P\left(\frac{1}{x^2}\leq U_1\right)^n = \left(\int_{\frac{1}{x^2}}^1 dy\right)^n = \left(1-\frac{1}{x^2}\right)^n$$ Note that this is for $x \geq 1$. For $x<1$, $P(M_n\leq x) = 0$.


0

NOTE: as the OP points out in the comments, this solution incorrectly assumes that ALL balls in $C$ are subsequently tossed over towards $D$. That case may still be relevant so I will leave this up for now and will modify it if I can incorporate the filter from $C$. Let $X_i$ denote the indicator variable for the $i^{th}$ red ball. Thus $X_i=1$ if $r_i$ ...


0

Compute the probability that the station will have to pay out for exactly one trip: $$\sum\limits_{n=1}^{1}\binom{35}{n}\cdot\left(\frac{1}{50}\right)^{n}\cdot\left(1-\frac{1}{50}\right)^{35-n}$$ Compute the probability that the station will have to pay out for more than $2$ trips: $$\sum\limits_{n=3}^{35}\binom{35}{n}\cdot\left(\frac{1}{50}\...


1

Suppose that 60% of a large population know how to spell the word 'accommodation'. The phrasing of this problem is poorly worded.   Let us instead suppose that $60\%$ of the large population will spell the word correctly when asked. As the population is large, then we may approximate that the probabilities of this happening for members of a small (...


0

What is the probability that exactly $2$ of the $5$ spelled the word correctly? $$\sum\limits_{n=2}^{2}\binom5n\cdot\left(\frac{60}{100}\right)^{n}\cdot\left(1-\frac{60}{100}\right)^{5-n}$$ What is the probability that at least one person spelled the word correctly? $$\sum\limits_{n=1}^{5}\binom5n\cdot\left(\frac{60}{100}\right)^{n}\cdot\left(1-\frac{...


0

Hint If $\varepsilon<1$, $$\left|-\frac{1}{\sqrt{\varepsilon}}\exp\left\{-\frac{(x-p)^2}{\varepsilon}\right\}\right|\leq \frac{\sqrt \varepsilon}{(x-p)^2}\leq \frac{1}{(x-p)^2}\in L^1(0,\infty ).$$ Therefore, convergence dominated theorem allow you to conclude.


0

could the unmatch part be calculated as (5!-(one match)-(two matches)-(3 matches)-(4 matches)-(5 matches))=5!-C(5,1)-C(5,2)-C(5,3)-C(5,4)-C(5,5)?


0

Let $X$ be the number of color matches that occur in a round of this game. Consider $P(X = 3)$. We know that for the event $\{X = 3\},$ we need three of the balls to match the color assigned to them. Hence, Given $8$ balls, there are $\binom{8}{3}$ ways to get three balls such that those three balls will match colors with their placed buckets. We now need to ...


0

The reasoning that led to the $\binom{8}{3}$ was good. We now need to multiply by the number of permutations of $5$ objects that leave no object fixed. This is the number of derangements of $5$ objects, and is often denoted by by $D_5$. So the number of favourables is $\binom{8}{3}D_5$. It turns out that $D_5=44$. And yes, then for the probability one ...


0

There is a slight misstep in notation often given with the conditional probability rule. In particular, $$P(X) = P(X|Y)P(Y)$$ intuitively assumes that we are conditioning upon the entire support of $Y$ ($S_Y$), and assuming $Y$ is a discrete random variable (the continuous case can dealt with similarly), we can see that the more appropriate notation would ...


1

$\vec z=\vec x\Vert\vec y~$ is the vector formed by concatenating vectors $\vec x$ and $\vec y$.   That is that: $$\vec x=\begin{pmatrix}a_1 \\ a_2\\ \vdots \\ a_n\end{pmatrix}, \vec y=\begin{pmatrix}b_1 \\ b_2\\ \vdots \\ b_m\end{pmatrix}\\ \vec z= \vec x\Vert \vec y = \begin{pmatrix}a_1 \\ a_2\\ \vdots \\ a_n\\b_1 \\\vdots \\ b_m\end{pmatrix}$$ Then ...


1

We see that $$P(Y|X) \cdot P(X) = \frac{P(Y,X)}{P(X)}\cdot P(X) = P(Y,X) = P(Z),$$ Where $Z$ is defined as the variable over the cartesian product of the supports of $Y$ and $X$ (as in, a realization $z$ of $Z$ is in $S_Y \times S_X$). Hence, 1. is the more appropriate answer.


0

I think I can provide a simpler answer: You know that the normal distribution is an even function so you know that the expectation of an odd power must be zero. You know the variance and mean of $z^2$ and you know that var($z^2$)= E[$(z^2)^2$] - E[$z^2$]$^2$ so you know then that E[$z^4$] = var($z^2$)+E[$Z^2$]$^2$ = 2+1=3


1

If you know a little bit of calculus, then there is a way to quickly calculate any moment you want for a normal distribution using its Moment Generating Function (MGF). In particular, the MGF for a normal distribution is: $$M(t)=e^{t\mu+\frac{1}{2}\sigma^2t^2}$$ The great property of this function is that $E(Z^k)=M^{(k)}(0)$. So, for your case where $\mu=...


1

Let $X$ be the number of attempts without a collision. Then, we have $\mathbb{E}[X]$ $= \displaystyle\sum_{k = 1}^{\infty}k\Pr[X = k]$ $= \displaystyle\sum_{k = 1}^{\infty}\sum_{n = 1}^{k}\Pr[X = k]$ $= \displaystyle\sum_{n = 1}^{\infty}\sum_{k = n}^{\infty}\Pr[X = k]$ $= \displaystyle\sum_{n = 1}^{\infty}\Pr[X \ge n]$. The probability that $X \ge n$ is ...


2

No, the calculation of the expected number of attempts before the first repeated value is a completely different problem than the number of attempts needed for the probability of a hit is some given $\rho$. The sum you are doing in the expected value problem is $$1 + \frac{1}{H} + 2 \frac{H-1}{H} \frac{2}{H} + 3 \frac{H-1}{H} \frac{H-2}{H} \frac{3}{H} + \...


1

For the loss of $k$ to kick in, he needs to win $k-a$ times. If he does that, he will never go broke (except maybe on the round he would quit because of the $k$ losses). He needs to win those $k-a$ within the first $2k-a$ games. So compute the chance he goes broke in less than $2k-a$ games and the expected length of a game in that scenario. This gives ...


1

The probability that $n$ distinct items are ordered in the favorite order after one minute is $\frac1{n!}$ The probability that $n$ distinct items are ordered in the favorite order after k minutes is $P(X=k)=\frac1{n!}\cdot \left(1-\frac1{n!} \right)^{k-1}$ The expected value is related to the geometric distribution. $$E(X)=\sum_{k=1}^{\infty} \frac1{...


1

Suppose he must sort the numbers $\{1,2,\dots,n\}$ If $\mu$ denotes the expected number of minutes then: $$\mu=\frac1{n!}1+(1-\frac1{n!})(1+\mu)$$ Here $\frac1{n!}$ is the probability that he is ready after one minute. If he is not then he must start over again and has lost one minute. This equality tells us that : $$\mu=n!$$ More formally: If $E$ ...


1

Assuming they're all distinct, the process succeeds with probability $1$ divided by the number of possible permutations, which is $n!$ for a sequence of $n$ positive integers. So it takes $n!$ steps on average.


2

$a$ is determined by the equation: $$\int_{-\infty}^\infty f(x) \ dx = 1$$So for this problem:$$a\int_{0}^\infty e^{-2x}-e^{-3x}\ dx = 1$$Now you can solve for $a$. For the second question, $$P(X\le 1)=\int_{0}^1 f(x) \ dx$$ Plug the values in and you can solve for the probability.


1

Hint: $a)$ to be a density function, it must be : $\displaystyle \int_{-\infty}^{\infty} f(x)dx = 1$. Split the integral. $b)$ $P(X\leq 1) = \displaystyle \int_{-\infty}^1 f(x)dx= \displaystyle \int_{0}^1 f(x)dx$


3

Assuming all the numbers are different, the probability is $\frac{1}{n!}$ of getting the right permutation in one try. Suppose $k$ of them are the same number, the probability is $\frac{k!}{n!}$ of getting the right permutation in one try. Suppose we have classes of sizes $k_1,\ldots,k_m$ such that each number in a class is equal. Then we have probability $\...


0

I'd agree with Clarinetist. Consider a simpler example: Define $$ x + f(x) = c, $$ where $c$ is some constant. We cannot simplify even this one to some general form. You can see that if we let $f(x) = x^{2}$, we have $$ x + x^{2} = c $$ $$ \Rightarrow x = \frac{-1\pm \sqrt{1+4c}}{2} $$ which we cannot (without knowing $f$) define as some inverse function ...


0

The derivation of the Normal Distribution (Gaussian Distribution) should answer your question; it also explains why there is a pi and so forth: https://www.sonoma.edu/users/w/wilsonst/papers/Normal/default.html


2

Since the pdf of the standard normal distribution is an even function, it follows that $\mathbb{E}[Z^3]=0$. And $$ \mathbb{E}[Z^4]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}x^4e^{-\frac{x^2}{2}}\;dx=\sqrt{\frac{2}{\pi}}\int_0^{\infty}x^4e^{-\frac{x^2}{2}}\;dx$$ which can be evaluated by setting $u=\frac{x^2}{2}$ and using the properties of the Gamma ...


0

In the second formula you give for the number of was to draw a two pair which is a product of $6$ terms, the outcomes (4, heart and spade, 3, heart and diamond, J, club) and (3, heart and diamond, 4, heart and spade, J, club) are counted as different hands when in fact they are the same five-card hand. So you need to divide this value by $2$ to get the ...


1

The question, at least in the way you asked it, really doesn't make much sense, because that funny exponential formula is the definition of "normal". Your comments sound like you sort of don't get what I'm getting at here. An analogy: Suppose someone asked this: Q: How do you prove that a triangle has three sides? The answer would be this: A: Huh? Having ...


0

We know for a continuous random variable, the PDF is the first derivative of the CDF wherever the CDF is differentiable. Here, as you see, the random variable is defined over the interval $ 1\le\ X \le\ 4 $ and the probability that it lies outside this interval is zero. Now, within this interval, the CDF is a piecewise function with only one point where ...


0

Firstly, to have a continuous function $F$, you need to solve these equations which should be satisfied on the boundaries $F(1)=0.2(1)^2-0.4(1)+b=0$ $F(2)=0.4(2)+a=0.2(2)^2-0.4(2)+b$ Then, either you can find the corresponding PDF or use another formula to find the expected value, which works for positive random variables $\int\limits_0^{\infty} (1-F(t))...


0

F(x)= 0 for t< 1 and $0.2t^2- 0.4t+ b$ for 1< t< 2. In order for the function to be continuous at t= 1, those must "match up": $F(1)= 0= 0.2(1^2)- 0.4(1)+ b= -0.2+ b$ so we must have $b= 0.2$. For 2< t< 4, F= 0.4t+ a. Again the two formulas must give the same value at t= 2. We must have $0=.2(2^2)- 0.4(2)+ 0.2= 0.4(2)+ a$. That reduces to ...


1

Hint: $P(X=1|X\geq 1)=\frac{P(X=1 \cap X\geq 1)}{P(X\geq 1)}=\frac{P(X=1) }{P(X\geq 1)}$ and $P(X\geq 1)=1-P(X=0)$


0

A sum of independent bernoulli random variables (which you've noted each have probability $\frac12$) will be a binomial random variable, but in this case more importantly the resulting sum's cumulative distribution function will tend to that of the normal distribution with the same mean and variance, by the central limit theorem. That should answer both ...


0

The event "Problem A is solved" gets the name $A$, and the event "Problem B is solved" gets the name $B$. Then the event "At least one problem is solved" can be renamed into "Problem A is solved OR problem B is solved" (remember that in logic and mathe matics, "or" means "either one or both"), which may be shortened (using our names) to "$A$ OR $B$". ...


0

By independence of $X$ and $Y$ we have $P(X>2) * P(Y>2) = P(X > 2 \wedge Y>2)$. Now let $A$ be the event $X>2 \wedge Y>2$ and $B$ the event $X+Y>4$. It is clear that if we have $A$ then we have $B$. The converse however is not true : $$ A \Rightarrow B \\ B \nRightarrow A $$ Therefore $P(B) \geq P(A) \Longleftrightarrow P(X+Y>4) \geq ...


0

As it happens, probability of sorted array with duplicate numbers was asked just a few hours ago. If number $x_j$ is present $n_j$ times, with $1\le j\le k$ and $\sum_jn_j=n$, the probability for a uniformly randomly chosen permutation to order the list is $$ \frac{\prod_jn_j!}{n!}\;, $$ so the expected number of permutations required is $$ \frac{n!}{\...


0

Comment. As an example, suppose $X, Y \sim Exp(rate=1)$ independently. Computation and plot in R for 30,000 simulated realizations. Based on hint by @NicholasStull. (Also, illustrates subsequent answer by @Zubzub.) x = rexp(30000); y = rexp(30000) mean(x > 2)*mean(y > 2) ## 0.01812118 # Aprx P(X > 2)*P(Y > 2) mean(x + y > 4) ## 0....


0

Take two poisson random variables $A$ and $B$ with means $\lambda_A$ and $\lambda_B$ respectively. We see that $$P(A > B) = \sum_{k = 0}^{\infty} P(A > B | B = k)P(B = k)$$ $$=\sum_{k=0}^{\infty} P(A \geq k + 1)P(B = k) = \sum_{k= 0}^{\infty} \left(\sum_{l=k+1}^{\infty} \frac{\lambda_A^{l} e^{-\lambda_A}}{l!} \right)\frac{\lambda_B^k e^{-\lambda_B}}{k!...


0

You are finding the average of 3 numbers so you divide the sum by 3.


0

$\Pr[R=i|R\ge 4]$ means "probability that comes out $i$ when only $4,5,6$ are possible outcomes". You have only $3$ possible outcomes so for each outcome $i$ the probability is $1/3$ if it is possible or $0$ if is impossible. Another way to see that is decomposing the conditional probability, i.e. $$\Pr[R=i|R\ge 4]=\frac{\Pr[R=i\text { and }R\ge 4]}{\Pr[R\...


0

The factor of $ \frac{1}{3} $ applies because that is the probability of rolling a 4, given the result is 4 or greater and the same for 5 and 6. Thus the expected value is 5.


1

Because the question asks for the excepted value of the face on a die given that the number rolled is at least $4$, we know for certain that the probability space only includes $4,5,6$. The faces on the die, $1,2,3$, occur with probability $0$ given that the number rolled is at least $4$. Since only $4,5,6$ occur and, in a fair die, each face would occur ...


0

The difference between a full house (AAABB) and a two pair (AABBC) is that when drawing a two pair, both pairs are groups of two. With the full house, one is a group of three, and the other is a group of two. With your methos, you are counting, for example, AABBC as different from BBAAC. So the logic is first choose the two (different) kinds from a total of $...


5

The problem with your $247,104$ is that it counts each two-pair hand two times, according to which of the pairs you mention first. But 5H-5D-7S-7H-9D is the same hand as 7S-7H-5H-5D-9D, so it gets counted both with fives first and with sevens first. In contrast, for a full house it is unambiguous which value is the one with three cards and which is the one ...


2

Usually, if the numbers were not repeated the probability would be 1/n! where n! is the number of possible sorted combination while just one is the correct one. As the number can be repeated, the number of each repeated numbers (for example {2,2,...,2}) can be between 1 to n and the number of different repeated numbers can be between 2 and n/2 (for example ...


6

I'll assume that you intended to imply that all permutations of the $n$ numbers are equiprobable. If number $x_j$ is present $n_j$ times, with $1\le j\le k$ and $\sum_jn_j=n$, there are $$ \binom n{n_1,\ldots,n_k}=\frac{n!}{\prod_jn_j!} $$ distinguishable arrangements. They are equiprobable and exactly one of them is ordered, so the probability for this ...



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