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30

The problem here is that your phrase "one gets head" is not precise enough. If it means "at least one of the coins comes up heads", then indeed there are 3 equally likely possibilities (HT,TH,HH) out of which exactly 1 has both coming up heads. This means that the desired probability is $\frac13$. If it means "out of the two coins A,B that were flipped, A ...


9

I think this is one of the cases where logic/mathematics totally goes overboard on a trivial problem, and you (as well as your friend) are overthinking it. But it's fun anyway, so... There are at least 3 possible answers which are equally correct, depending on how pedantically one tries to twist the wording one way or the other. But for practical purposes ...


6

Probability is $3/4$ if points $x$ and $y$ are chosen with uniform probability. That corresponds to the area in color in the picture below.


5

With probability $\frac 12$ both points are on the same side of the midpoint, so we are guaranteed success. If the points are on opposite sides of the midpoint(a probability $\frac 12$ event, with $P<\frac 12< Q$ say, then again with probability $\frac 12$ we have $Q$ is nearer $1$ than $P$ is near $\frac 12$,so the segment between them has length ...


5

\begin{align*} P(X>Y) &= \int_{0}^\infty\int_{0}^x abe^{-ax}e^{-by}\,dydx\\ &=\int_0^\infty ae^{-ax}\left[\int_0^xbe^{-by}\,dy\right]\,dx\tag 1\\ &=\int_0^\infty ae^{-ax}\left(1-e^{-bx}\right)\,dx\\ &=\int_0^\infty ae^{-ax}-a e^{-(a+b)x}\,dx\tag 2\\ &=1-\frac{a}{a+b}\int_0^\infty(a+b)e^{-(a+b)x}\,dx\tag 3\\ ...


4

It's called conditional probability: Let $A$ denote the event of getting exactly $2$ heads. Let $B$ denote the event of getting at least $1$ head. Then the probability that $A$ will occur given that $B$ has occurred is: $$\frac{P(A\cap B)}{P(B)}=\frac{1/4}{3/4}=\frac13$$ EDIT: Please note that in this specific case, $A\cap B=A$. Therefore, one could ...


3

Assuming you sent the resumes off to companies entirely at random (without looking where they were based, in particular) then: There are $\binom {14}{6}$ ways in which you might have sent them off. There are $\binom {11}{6}$ ways to send them only to the capital city firms. Your answer is just the ratio: $$\frac {\binom {11}{6}}{\binom {14}{6}}\sim ...


3

$$ \frac{1+\xi}{3-\xi}=\frac13\frac{1+\xi}{1-\frac\xi3}=\frac13(1+\xi)\sum_{k=0}^\infty\left(\frac\xi3\right)^k\;. $$


3

Recall from conditional probability that $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$. Here, your events are knowing the answer and answering correctly respectively. You are already given the probability of knowing the answer and answering correctly, precisely because the event of knowing the answer is a subset of answering correctly (if you know the answer, you ...


3

Hint: We will consider the situation in terms of minutes. We notice that that they will meet only if $|X-Y|\leq 5/60$, where $X,Y$ is the time of arrival as a fraction of the hour. We notice that $X,Y\overset{iid}{\sim} \text{unif(0,1)}$. Then they problem becomes $$P(|X-Y| \leq 5/60).$$ It will help to draw a picture. No integration required. You can do ...


3

You can skip that step and instead just substitute $u=3y$ immediately into $$\frac{2^x}{x!}\int_0^\infty e^{-3y}y^xdy$$ to get $$\frac{2^x}{x!}\int_0^\infty e^{-u}(u/3)^x\frac{du}{3}=\frac{2^x}{x!}\int_0^\infty \frac{e^{-u}u^xdu}{3^x\cdot 3}=\frac{2^x}{x!3^{x+1}}\int_0^\infty e^{-u}u^xdu.$$ This of course gives the same result, but doesn't require you to ...


3

This kind of trick is often clear in retrospect and not always easy to see immediately. Practice surely helps! The more problems you solve the easier it will be to see a few steps ahead. In this case, notice that the integrand in last equation on the first line is $e^{-3y}y^x$. This equation sort of looks like the expectation of $y^x$ where $y$ is an ...


2

Think of it this way: If you toss the first coin until it comes head and only then toss the second, the probability of the second being head is 1/2. Because it is independent of the first coin. Now, toss the first coin again and no matter the result, toss the second coin. Obviously, there is a difference. You do not know beforehand that you will get a head ...


2

Yes the reasoning is correct. You can solve the problem by assuming that the coins are distinguishable, and you get the following counts: $$\begin{align}HH&: 1,\\ HT&: 1,\\ TH&: 1,\\ TT&:1.\end{align}$$ This is a uniform distribution. Then the requested probability is $$\frac{HH}{HH+HT+TH}=\frac13.$$ Or you can consider that they are ...


2

Hint: If $f_{X,Y}$ is the multivariate pdf then you want to solve the following integral $$ \int_{0}^{60}\int_{\max\{0,x-5\}}^{\min\{60,x+5\}} f_{X,Y}(x,y)dydx. $$


2

There is an alternative approach see this reference by first determining the pdf of $X-Y$ (or find it in tables) $$f(x) = \frac{ab}{a+b} \begin{cases}e^{b x} & \text{if} \ x \leq 0\\ e^{-a x} & \text{if} \ x \geq 0 \end{cases} \ \ \ (*)$$ Remarks: 1) the curve of $f$ is tent-shaped (see figure below for the case $a=3$ and $b=1$). 2) a ...


2

Your concrete problem is already solved here: If we throw three dice. The general problem is equivalent to counting the number of ways of distributing $X-Y$ balls into $Y$ bins with limited capacity $Z-1$. This problem is solved at Balls In Bins With Limited Capacity using inclusion-exclusion. The result is $$ \sum_{t=0}^Y(-1)^t\binom ...


2

In the actual solution, they choose Z as the first pick, X as the second pick, and Y as the third pick. What?   No, we just applied the Law of Total Probability and the definition of conditional probability.   These are general rules; they always apply. $$\begin{align} \mathsf P(Y\mid Z) =&~ \sum_{X\in\mathcal X}\mathsf P(X,Y\mid Z) ...


2

Ok suppose you take some point $x\in[0,1/2]$. Now taking a second point $y\in[0,1]$, there are two situations where you obtain a segment of length at least $1/2$. Firstly if $y\le 1/2$. Secondly if $y\ge x+1/2$. So for taking a first point $x$, the chance that you have a segment of desired length is $1/2+(1-1/2-x)=1/2+x$. Now integrating this over $[0,1/2]$ ...


2

I have three events, $C$ is the event that the problem was correct, $K$ is the event that the student knew the answer, and $G$ is the event that the student guessed. Now, we proceed mechanically, and the problem becomes \begin{align*} P(K|C)&= \frac{P(K,C)}{P(C)}\\ &=\frac{P(C|K)P(K)}{P(CG)+P(C\bar G)}\\ &=\frac{P(C|K)P(K)}{P(C|G)P(G)+P(C|\bar ...


2

Actually your teacher explained it pretty well. It is a trick that makes integral calculation easier. Your teacher is trying to match $e^{-3y}$ to $3^xy^x=(3y)^x$ in order to use the substitution method $u=3y$ Because $\frac{3^x}{3^x}=1$, multiplying this term to the whole integral will not change the result. If you have something like $e^{-6y}$ instead in ...


1

As you know how to integrate $e^{-y}y^xdy$, you are trying to make the integrand look like that function. You already have $e^{-3y}y^xdy$, and $3^x$ is a constant (w.r.t. $y$). So, in order to make the integrand look like what you want, you expand the integrand by $3^x$, then divide the whole thing by the same number to even things out.


1

\begin{eqnarray} P [ X+Y \le \alpha] &=&\int_{x=-\infty}^\infty \int_{y=-\infty}^\infty 1_{\{(x,y) | x+y \le \alpha \}} ((x,y)) f_X(x) f_Y(y)dy dx \\ &=& \int_{x=-\infty}^\infty \int_{y=-\infty}^{\alpha-x} f_X(x) f_Y(y) dy dx \\ &=& \int_{x=-\infty}^\infty \int_{y=-\infty}^{\alpha} f_X(x) f_Y(y-x) dy dx \\ &=& ...


1

You cannot unless you know how the weights for the second card are affected by extracting the first. IE: the conditional probabilities. $$\mathsf P(\{X_1,X_2\}=\{x,y\}) ~=~ \mathsf P(X_1=x)~\mathsf P(X_2=y\mid X_1=x)+\mathsf P(X_1=y)~\mathsf P(X_2=x\mid X_1=y)$$


1

It is because $0\leq y\leq x$ is the support if $Y$. When you factor in the support of $X$, the support of the joint distribution is , $0\leq y, \max(1,y)\leq x$ $$\begin{align}f_{Y}(y) =&~ \int_{y}^\infty f_{X,Y}(x,y)\operatorname d x~\big[0\leq y\big]\\[2ex] =&~ \int_{\max(1,y)}^\infty \tfrac 1{x^3}\operatorname d x~\big[0\leq ...


1

In the general case, neither of these is right. You have two inequalities for $x$ in $f(x,y)$, namely $x\ge1$ and $x\ge y$, so the lower limit of the integral is $\max(1,y)$. You appear to be implying that the lower limit $y$ was specified in some book or lecture or the like. If so, I suspect that this was done because you need the value for $Y=\frac32$, and ...


1

Assuming $\delta=1-\epsilon\ge0$, for $x\ge0$ we have \begin{align} \def\pro#1{\textsf{Pr}\left(#1\right)} \pro{X\ge x}&=\pro{R\cos\Theta\ge x}\\ &=\pro{\cos\Theta\ge\frac xR}\\ &=\int_\delta^1\pro{\cos\Theta\ge\frac xr}\mathrm dr\\ &=\begin{cases} \frac1{\epsilon\pi}\int_\delta^1\arccos\frac xr\mathrm dr&0\le x\le\delta\\ ...


1

This is called a derangement. It's hard to solve numerically exactly. From http://en.wikipedia.org/wiki/Derangement as n approaches infinity, permutation(n) divided by derangement(n) is 1/e


1

Hint: If you cut the circle along the first placed point, you can see that the situation is equivalent to taking the interval $[0,1]$ and placing $n-1$ points uniformly at random into the interval.


1

The probability for $j$ particular variables to be at least $\frac1n$ is $$ \left(1-\frac jn\right)^{n-1}\;, $$ so by inclusion-exclusion the probability for exactly $k$ variables to be at least $\frac1n$ is $$ \sum_{j=k}^n(-1)^{j-k}\binom nj\binom jk\left(1-\frac jn\right)^{n-1}\;. $$ I'm not aware of any way to simplify this.



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