Recently I read about how many people it takes for the probability that two or more share the same birthday to be sufficiently high.   My question: How many people do I have to know that I could be sufficiently sure (>95%) that I have to congratulate at least one person every day for the whole year, that is, all 365 days?
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Suppose Earth has been attacked by $n$ Martian spaceships and suppose that we have $m=k \cdot n$ missiles to release against the $n$ spaceships. The probability to hit and destroy each spaceship by each missile is $p$ (independent from the rest).  What is the probability to destroy all the spaceships if we release all the missiles at the same time but each missile will choose a spaceship at random?
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The question itself is a very easy one:     Somebody has got two kids, one of whom is a girl. Then what's the probability that he's got at least one boy?   My answer is that, since he's already got a girl, then "he's got at least one boy" amounts to "the other kid is a boy", whose probability is apparently $\frac{1}{2}$. But my friends argue that the probability should be $\frac23$: they say this is a binomial distribution, all the possible cases are (girl,girl),(girl,boy),(boy,girl) which yields that the probability is two cases out of three and is thus $\frac23$. But I think this is totally unacceptable. I don't think it is a binomial distribution at all, at least not what my friends explained to me. However, I just can't disuade them of their opinion, nor can I prove that I am wrong. So what on earth is the probability? and why? Any help is appreciated. Thanks in advance. Esp. Can anybody show why my explanation is wrong? Isn't it that whether the other kid is a boy or a girl a 50/50 event?  EDIT: Thanks for all the help you provided for me, and special thanks will go to @HammyTheGreek and @KSmarts, who have made it clear to me that there is in fact some ambiguity in my statement in this problem. As is pointed in  ,two distinct interpretation of the statement "one of whom is a girl" that gives rise to ambiguity:     From all families with two children, at least one of whom is a boy, a family is chosen at random. This would yield the answer of 1/3.   From all families with two children, one child is selected at random, and the sex of that child is specified to be a boy. This would yield an answer of 1/2.
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"An event less frequent in the long run is likely to happen before a more frequent event!"  How can I show that THTH is more likely to turn up before HTHH with a probability of  9/14, even though waiting time of THTH is 20 and HTHH, 18!  I would be very thankful if you could show me the way of  calculating the probability of turning up earlier,   and the waiting time. Thank you!
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How would you solve this probability problem?     Group of men are playing game. Each rolls a dice.      If the number is 1, he must say "I like pink" regardless of his true taste and       If the number is 2, he must say " I don't like pink" regardless of his true taste.       If the number is 3,4,5 or 6, he must tell his true taste.    Assuming that 2/3 of men answered "I like pink", what is the probability that a man actually likes pink?
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I heard an interesting question recently:     What is the minimum number of people required to make it more likely than not that all 365 possible birthdays are covered?   Monte Carlo simulation suggests 2287 ($\pm 1$, I think).  More generally, with $p$ people, what is the probability that for each of the 365 days of the year, there is at least one person in the group with that birthday?  (Yes, ignoring the leap-day.)
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I'm trying to figure out what is the probability with 3 D10's if I were to roll them one after another."001" is one and "000" is a thousand. Would every number have the same probability or doubles and triples be less common?  And please explain.
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If I am selecting 232 people from a pool of 363 people without replacement what is the probability of 2 of a list of 12 specific people being in that selection?  This is a random draw for an ultra race where there were 363 entrants for 232 spots. There is an argument about whether the selection was biased against a certain group of 12 people.  My initial attempt at calculating this was that there was 232 choose 363 possible selections. The number of combinations of any one person from the list of twelve is 1 choose 12 + 2 choose 12 + ... + 11 choose 12 + 12 choose 12. Thus 1 choose 12 + 2 choose 12 .... / 232 choose 363. Which ends up being a very low number which is clearly too low.  How do I calculate this?
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Given a box that contains 90% fair coins and 10% loaded coins, (a loaded coin gives heads 90% of the time), what is the probability for a randomly drawn coin to give 5 heads in a row?  P(HHHHH | random coin from box of mixed coins) = ?   I used Bayes Theorem to calculate the probability I got .219 or 21.9% probability for the random coin to give 5 heads in a row!    But I am still doubting if my answer is correct !!!  Any help appreciated :)
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Question:  Consider a group of n people with the following properties:  • no person is friends with everyone, • any pair of strangers share exactly one friend in common,  • no three people are mutually friends. Show that everyone has the same number of friends.  I want to solve this using Ramsey's theorem but I'm struggling to formulate it in a way that would make it straightforward.. Any help would be greatly appreciated.
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I gave my friend  as a brainteaser; while her attempted solution didn't work, it raised an interesting question.  I flip a fair coin repeatedly and record the results.  I stop as soon as the number of heads is equal to twice the number of tails (for example, I will stop after seeing HHT or THTHHH or TTTHHHHHH).  What's the probability that I never stop?  I've tried to just compute the answer directly, but the terms got ugly pretty quickly.  I'm hoping for a hint towards a slick solution, but I will keep trying to brute force an answer in the meantime.
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I confirmed on  thread that there are as many as even natural numbers as there are natural numbers.   Question : Suppose I have selected a number $n \in \mathbb N$; what is the probability that $n$ is even?  My Thought : $\text{Probability} = \dfrac{\text{n(E)}}{\text{n(S)}}$ Here $\text{n(S)}$ is the set of all natural numbers i.e. $\mathbb N$, and $\text{n(E)}$ is set of all even natural numbers. Since it is proved that number of elements is the set $\mathbb N$ is exactly the same as the number of elements in the set of natural numbers (it’s very easy to put the set of natural numbers, $\Bbb N=\{0,1,2,3,\dots\}$, into one-to-one correspondence with the set $\text{E}=\{0,2,4,6,\dots\}$ of even natural numbers; the map $\Bbb N\to \text{E}:n\mapsto 2n$ is clearly a bijection.) ; Thus, Probability $= \boxed 1$ I know this is definitely wrong.Probability must be $0.5$. But where am I wrong? Can anyone explain ? Thanks!
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There are $2014^{2014}$ children at a mathematics camp. Each has at most three friends at the camp, and if A is friends with B, then B is friends with A. The camp leader would like to line the children up so that there are at most $2014$ children between any pair of friends. Is it always possible to do this?  I thought maybe the pigeon hole principle. Obviously the 2014 is arbitrary. I googled it looking for a solution and it turns out it is from the BMO round 2. there it is 2010. Any help would be much appreciated.
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Problem:     Two people are to arrive at the same location some time between 3pm   and 4pm. They both arrive at random times within the hour and only   stay for ten minutes. Since they only go one time, what is the   probability that they will meet during that hour?   I have found the correct answer to this problem here (the second formula P2):      $$ P2 = 1 - \frac{(1-w_1)^2 - (1-w_2)^2}{2} = w_1 + w_2 - \frac{w_1^2}{2} - \frac{w_2^2}{2} $$  Plugging in ($1/6$) for both $w_1$ and $w_2$ is $11/36$ which is the correct answer.  But I don't understand how they came to that conclusion and what formula they derived it from. Any help in explaining this would be much appreciated!
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Given there are 365 days in a year and $n$ people, the probability that at least two people have the same birthday is given by  $$\frac{365^n - 365 \cdot 364 \cdots (365-n+1)}{365^n}.$$ In this calculation, we treat the $n$ birthdays we selected as a sequence where their order matters.   If we think the $n$ dates that randomly selected as a set where the order does not matter, I believe we would have the probability $$\frac{{365+n-1 \choose n} - {365\choose n}}{{365+n-1 \choose n}}.$$   For the total number of outcomes, we model by throwing $n$ balls uniformly randomly into $365$ boxes. Essentially, there are $n$ balls and $365-1$ walls (between the boxes). For example $$ \circ |\;\;\; |\;\;\;|\circ \circ |\;\;\;|\circ |\cdots |$$ would mean one person was born on day $1$, two people were born on day $4$, one person was born on day $6$, and so on. This is how I came up with the ${365+n-1\choose n}$. For the case that everyone has a distinct birthday, it would be ${365 \choose n}$.   What is wrong with this approach?
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There are many descriptions of the "birthday problem" on this site — the problem of finding the probability that in a group of $n$ people there will be any (= at least 2) sharing a birthday.  I am wondering how to find instead the expected number of people sharing a birthday in a group of $n$ people. I remember that expectation means the weighted sum of the probabilities of each outcome:  $$E[X]=\sum_{i=0}^{n-1}x_ip_i$$  And here $x$ must mean the number of collisions involving $i+1$ people, which is $n\choose i$. All $n$ people born on different days means no collisions, $i=0$; two people born on the same day means $n$ collisions, $i=1$; all $n$ people born on the same day means $n$ collisions, $i=n-1$.  Since the probabilities of three or more people with the same birthday are vanishingly small compared to two people with the same birthday, and decreases faster than $x$ increases, is it correct to say that this expectation can be approximated by  $$E[X]\approx {n\choose 0}p_{no\ collisions}+{n\choose 1}p_{one\ collision}$$  This doesn't look right to me and I'd appreciate some guidance.    Sorry - edited to change ${n\choose 1}$ to ${n\choose 0}$ in second equation. Sloppy of me.
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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?
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I am just learning probability in my Discrete Structures class and am very lost. This is the example given in the book and I have no idea how to solve this problem.   Problem: Suppose one in 1000 people have a certain disease. Suppose medical testing is not perfect (as is the case in real life) and consequently, only 99% of the people with the disease are tested positive. Suppose 2% of the people who don’t have the disease also test positive. What is the probability of actually having the disease, given someone tests positive?  Things I do know:   People with disease = 1/1000 Tested positive with disease = 99/100 People tested positive who don't have disease = 2/100   I'm not sure where to got from here with the information that I know. What are the next steps?