We have the unit interval $[0,1]$ and we want to find the probability of picking two random numbers $a,b$ from that interval with $|a-b|&gt;0.5$.  Must I investigate $[0,1]×[0,1]$?  I don't have the faintest idea of how to solve this. The problem is that $[0,1]$ has infinite numbers to pick from… so how to calculate a probability with infinitely many items in the sample space?  I would be really happy if somebody shed a light on this.
86065
Two numbers are chosen randomly and independently from the unit interval $(0,1)$.  What is the probability that they differ by more than $\frac{1}{2}$? Hint: Think of the $2$ numbers chosen as coordinates in the Cartesian plane. Draw a square with vertices at $(0,0)$,$(1,0)$,$(1,1)$, and $(0,1)$. Allowable choices for the $2$ points lie below the line $y=x-\frac{1}{2}$ or the above $y=x+\frac{1}{2}$. Find the combined area of these regions.  I need help on this. I know what $y=x-\frac{1}{2}$ and $y=x+\frac{1}{2}$ looks like but I am not understanding the problem as a whole
65014
A man visits a couple who have two children.One of them, a boy, comes in to the room. Find the probability p that other is also a boy  a) 1/3 b) 2/3 c) 1/2 d) 3/4  Correct Ans: a) 1/3  However I am getting answer as 1/2.My solution is as follows.  The couple has two children, hence the sample space is as follows.  B- Boy G - Girl  BB BG GG ( BG and GB mean the same thing as Girl,Boy), but it is already known that one is boy.So the sample space reduces to  BB BG  So now p is 1/2  Is there some mistake is my approach ?
98717
If two points are selected on a straight line of length 'a' units at random, then what is the probability that none of the three line segments formed by the two random points has length less than a/4.
64775
I'm working on some problems from Pitman's Probability.  Let events $A$ and $B$ be independent, with $P(A)=0.1$ and $P(B)=0.3$. Find: $(a)$ Neither of the events occur $(b)$ At least one of the events occur $(c)$ Exactly one of the events occur  Attempt: $(a)$ If neither of the events occur, this is the complement of both events occurring. Hence $1-P(A\cap B)=1-P(A)P(B)=1-\frac{1}{30}=\frac{29}{30}$ $(b)$ At least one of the events can be represented as the union of the two events. Hence $P(A\cup B)=P(A)+P(B)-P(A\cap B)=P(A)+P(B)-P(A)P(B)=\frac{4}{10}-\frac{1}{30}=\frac{11}{30}.$ $(c)$ Not sure what this would be. Maybe $P(A^c\;\cap\;B)+P(A\cap B)$?
62763
I was wondering if someone could critique my argument here. The problem is to find the probability where exactly 2 people in a room full of 23 people share the same birthday.  My argument is that there are 23 choose 2 ways times $\displaystyle \frac{1}{365^{2}}$ for 2 people to share the same birthday. But, we also have to consider the case involving 21 people who don't share the same birthday. This is just 365 permute 21 times $\displaystyle \frac{1}{365^{21}}$. To summarize:  $$\binom{23}{2} \frac{1}{365^2} \frac{1}{365^{21}} P\binom{365}{21}$$
7355
I found this math "problem" on the internet, and I'm wondering if it has an answer:     Question: If you choose an answer to this question at random, what is the probability that you will be correct?      a. $25\%$      b. $50\%$      c. $0\%$      d. $25\%$   Does this question have a correct answer?
52378
I am trying to solve interesting probability problem which I found in the book "50 Mathematical Ideas You Really Need to Know" by Tony Crilly:  A group of people go to lunch and afterwards pick up their hats at random. What is the probability that no one gets their own hat?"  So he claimed that It can be shown that this probability is $1/e$ (about $37\%$) so that the probability of at least one person getting their own hat is $1 – 1/e$ $(63\%$)  I was thinking about asymptotic cases where $n \to \infty$. The probabilities of "no one gets their own hat" and "at least one person getting their own hat" should be $0$ and $1$ respectively, right? So why it is $1/e$ and $1-1/e$?  Thank you in advance.
17348
I have this question that I don't know where to begin, any help will be greatly appreciated.  Consider the following random experiment. Toss a coin until the same number of heads and tails have been observed. Take as random variable, X, the observed number of heads (or tails). Determine the first few values in the probability distribution and calculate  $\sum^{5}_{n=1} nP (X = n)$  I'm not sure exactly where to start, if someone could point me in the right direction that would be great! thanks
86055
Say you have a stick which breaks randomly into three pieces (we can choose the points randomly). What is the probability that the area is greater than or equal to $0.4$?   I can see it has something to do with Heron's formula but I just can't put t together.
39932
Given $n$ items with weight $w_n$ each -- what is the probability that item $i$ is chosen in a $k$-out-of-$n$ "weighted random sampling without replacement" experiment? Can a closed-form solution that depends only on $w_i / w_\cdot$ be derived ($w_\cdot = \sum_j w_j$.)?   EDIT: A solution that depends only on $w_i / w_\cdot$ is impossible. Assume $n=3$, $i=1$, $w_1 = 1$, and two cases: (1) $w_2 = 1, w_3 = 1 + \varepsilon$, (2) $w_2 = 2, w_3 = \varepsilon &gt; 0$. In case (1) the probability is almost $2/3$, in case (2) it is almost $1$, but in both cases $w_1 = 1$ and $w_\cdot = 3 + \varepsilon$.  What I have tried so far to solve the problem:  Let $P^n_k(w, i)$ be the probability that item $i$ is chosen in an $k$-out-of-$n$ experiment with weight vector $w$. In the first draw, the item is chosen with probability $w_i / w_\cdot$. Otherwise, we are looking for the probability to choose this item in a $k-1$-out-of-$n-1$ experiment, with the same weight vector except for the item that has been selected. Hence:  $$ P^n_k(w, i) = w_i / w_\cdot + \sum_{j \neq i} w_j / w_\cdot \cdot P^{n-1}_{k-1}(w - j, i) $$ $$ = w_\cdot^{-1} \left( w_i + \sum_{j \neq i} w_j \cdot P^{n-1}_{k-1}(w - j, i) \right) $$  with $w-j$ being "the vector $w$ without the $j$th element".  How to solve this recurrence relation? (If it is correct at all...)
95556
The following is a question I've made myself, but I need help in solving it:  Suppose there are 100 balls in a box. 20 balls are blue, 30 balls are green and 50 balls are yellow. Now we randomly pick out 10 balls out of the box (one ball after the other) and we don't put the balls back in the box.  What's the probability of the 10th ball being picked having color blue?  I tried thinking of all the possibilities of the the 10th ball being blue, divided by all of the possible combinations. I tried for several hours and couldn't figure out neither of them. Help is appreciated!
69401
Suppose we uniformly and randomly select a permutation from the 20! Permutations of 1,2,3,...,20. What is the probability that 2 appears at an earlier position than any other even number in the selected permutation?  My approach :   2 in first position ==> 1*19! (1 is the position of 2)  2 in second position ==> 10 *1 *18!  2 in third position ==> 10*9 *1 *17!  . . .  2 in 11th position ==> 10*9*8*7*6*5*4*3*2*1*1*9!  But this was asked as a multiple choice question in GATE 2007 and the options were  (A) 1/2   (B) 1/10  (C) 9!/20!  (D) None of these  I am not able to reduce this analysis to a given option.
23075
How do I generate numbers based on an arbitrary discrete distribution?  For example, I have a set of numbers that I want to generate. Say they are labelled from 1-3 as follows.  1: 4%, 2: 50%, 3: 46%  Basically, the percentages are probabilities that they will appear in the output from the random number generator. I have a pesudorandom number generator that will generate a uniform distribution in the interval [0, 1]. Is there any way of doing this?   There are no bounds on how many elements I can have, but the % will add up to 100%.
62327
On googling I got     Random numbers are numbers that occur in a sequence such that two conditions are met: (1) the values are uniformly distributed over a defined interval or set, and (2) it is impossible to predict future values based on past or present ones. Random numbers are important in statistical analysis and probability theory.   What if I want to pick a random integer from the number line ? The number line consists of infinitely many integers. My favorable event is to get a number (random) and the number of events possible is the complete number line i.e. infinity. I won't be considering any limits here as our number line doesn't tend to infinity but does contain infinitely many elements.  Now the probability of getting a random number  P(E) = 1 / ∞ = 0   So I won't be able to get a random number, ever. This implies, never would I ever be able to get a random number.  That's absurd as many of our science topics do depends on things being random.
68945
I was implementing a simple scheme of Bernoulli distribution sampler. $ X \sim B(p) $. I have a function that generates a uniform random number $r \in (0,1)$. Then, I set $ X = 1 $  if $p &gt; r $, and $X =0$ otherwise. Is this correct?
111722
Banach Matchbox Problem:   Suppose a mathematician carries two matchboxes at all times: one in   his left pocket and one in his right. Each time he needs a match, he   is equally likely to take it from either pocket. Suppose he reaches   into his pocket and discovers for the first time that the box picked   is empty. If it is assumed that each of the matchboxes originally   contained $n$ matches, what is the probability that there are exactly $k$   matches in the other box?   I'm wondering whether the following reasoning is right, because it doesn't match up with the correct probability. But here goes my reasoning:  Assuming there are $k$ matches left in the other box, we have had to take $2n-k+1$ matches to notice there are none left. The total number of ways in which we could have picked those is $2 {2n - k \choose n}$, for either the matchbox in the left or right pocket has $k$ matches left inside, and in the $(2n-k+1)$th pick we would have found an empty box.   The total number of possibilities, taken over all possible sizes $k$, would then be $\sum_{m=0}^{n} 2{2n-m \choose n}$. So I'd assume the overall probability would be $\frac{2 {2n - k \choose n}}{\sum_{m=0}^{n} 2 {2n-m \choose n}} =\frac{{2n - k \choose n}}{\sum_{m=0}^{n} {2n-m \choose n}}$.  However, the mentioned solution is ${2n - k \choose n} (\tfrac{1}{2})^{2n-k}$. Where is my reasoning wrong? Thanks in advance.
84038
Given $20$ people, what is the probability that, among the $12$ months   in the year, there are $4$ months containing exactly $2$ birthdays and   $4$ months containing exactly $3$ birthdays?   I am less concerned about knowing the answer than making sure I have thought about the problem correctly.   The general problem seems to be finding the distinct integer-valued vectors that satisfy $$x_1 + x_2 + ... x_{12} = 20\ \ \ \ \ \ \ \ \ x_i \ge 0, i = 1, ..., 12$$  It can be proven that there are $\binom{20+12-1}{12-1}$ such vectors. I am presuming this to be the sample space.  (Please correct me if I'm wrong)  $4\cdot 2 + 4\cdot 3 + 4 \cdot 0= 20$, so there are $3$ distinct groups of $4$ months. Therefore we have $\binom{12}{4\ 4\ 4}$ possible ways to get the desired distribution of months.   The probability I get is $$\frac{\binom{12}{4\ 4\ 4}}{\binom{20+12-1}{12-1}} $$. The answer in the back of the textbook is $1.0604 \times 10^{-3}$ so my answer is off. I just want to know where I went wrong.