There is some  now about a Polish village where the last 12 children born were all girls.  The probability that the last 12 newborns in a village are all girls is around 0.03%, so around 3000 villages give us a solid 50% chance of this happening, so I would say this is no big deal.  However I was thinking how often this happens in large hospitals and goes unnoticed.   According to some  the Northside Hospital in Atlanta has seen 27,123 births in 2016.  So what is the probability that at least 12 girls in a row were born there in 2016?
124011
So a run of $x$ heads has the probability $0.5^x$.  Thefore a run of $9$ would be $0.5^9 = 0.001953125$.  But what is the chance of that run of $9$ occuring in $1000$ coinflips? Can anyone please explain how to work this out?
73535
So, I have been presented with this question: A sample of 100 people were asked how many days they drove their car during the last week (inc. the weekend). The resulting frequency of response is shown below: Days, frequency 0, 1 1, 5 2, 3 3, 15 4, 20 5, 25 6, 31 7, 0 I need to use a beta-binomial model in order to estimate the average number of 'car driven' days a week for a uniform prior. There are a total of 7 Bernoulli trials, each with success probability p. I'm incredibly confused at what the outcome will be. I know that for a uniform prior that alpha = 1, and beta = 1, but I don't know how to apply this all to get the average? Thank you.
64910
I'm trying to wrap my head around the result of Bayes Theorem applied to the classic mammogram example, with the twist of the mammogram being perfect.  That is,  Incidence of cancer: $.01$  Probability of a positive mammogram, given the patient has cancer: $1$  Probability of a positive mammogram, given the patient does not have cancer: $.01$  By Bayes:  P(cancer | mammogram+) = $\dfrac {1 \cdot .01}{(1 \cdot .01) + (.091 \cdot .99)}$  $ = .5025$  So, if a random person from the population takes the mammogram, and obtains a positive result, there is a 50% chance they have cancer? I'm failing to intuitively understand how the tiny 1% chance of a false positive in 1% of the population can trigger a 50% result. Logically, I would think a perfectly true positive mammogram with a tiny false positive rate would be much more accurate.
20556
My professor assigned a completely random question on our problem set. Basically, it goes:  Sophie is 30 years old and majored in philosophy. As a student, she participated in anti-nuclear demonstrations and was deeply concerned with social justice. Which is more probable?  1. Sophie is a bank teller. 2. Sophie is a bank teller and is active in the feminist movement.  I'm guessing the point the prof was trying to get across is that assuming you have two different events $A$ and $B$,  $A\cap B$ would always have a smaller probability than either. Is this correct?
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 ?
108025
If nine coins are tossed, what is the probability that the number of heads is even?  So there can either be 0 heads, 2 heads, 4 heads, 6 heads, or 8 heads.  We have $n = 9$ trials, find the probability of each $k$ for $k = 0, 2, 4, 6, 8$  $n = 9, k = 0$  $$\binom{9}{0}\bigg(\frac{1}{2}\bigg)^0\bigg(\frac{1}{2}\bigg)^{9}$$   $n = 9, k = 2$  $$\binom{9}{2}\bigg(\frac{1}{2}\bigg)^2\bigg(\frac{1}{2}\bigg)^{7}$$   $n = 9, k = 4$ $$\binom{9}{4}\bigg(\frac{1}{2}\bigg)^4\bigg(\frac{1}{2}\bigg)^{5}$$  $n = 9, k = 6$  $$\binom{9}{6}\bigg(\frac{1}{2}\bigg)^6\bigg(\frac{1}{2}\bigg)^{3}$$  $n = 9, k = 8$  $$\binom{9}{8}\bigg(\frac{1}{2}\bigg)^8\bigg(\frac{1}{2}\bigg)^{1}$$  Add all of these up:   $$=.64$$ so there's a 64% chance of probability?
30856
I share the same birthdate as my boyfriend, same date but also same year, our births are seperated by merely 5 hours or so.  I know that the chances of meeting someone who was born on the same date than me is fairly high and I know a few people with whom I share my birthday although for the little I've read about the birthday paradox, it doesn't take same year into account.  We've argued before about the probabilities and I am still not satisfied. My point was that the chances are tiny if you consider the probabilities of being in a relationship (+ being successful at it for X amount of time). I find the amount of factors to take into account quite vast (up to a point, gender and age, availability, probabilities of separation in our region, etc.)  Is it even possible to calculate the probabilities on something like this? How would you go about it?
59448
The question is: n balls are distributed into n holes at random. What is the probability that exactly one hole remains empty. I came up with $$P\left(A\right)=\frac{\:\dbinom{n\:}{1}\:\dbinom{n\:-1}{1}\:\left(n-2\right)!}{n^n}$$ but was told I’m way way off. Yes, I do realize I’m awful at probability. Thanks for the help.
103156
I have no idea what the first problem on this list is asking? The others I can figure our, but does anyone know what the first one is actually asking? If so, how would you set it up?  Nate can only invite 8 people to his birthday, but has 13 friends he’d like to invite.  Eight of his friends are boys and five are girls.  Three of the girls are roommates and must stay together.  If Dante randomly selects 8 people, determine the probability for each scenario.   Dante’s random selection is a valid outcome Dante does not invite any girls Dante invites an equal number of boys and girls
26894
There are $200$ students in a class. Each student is a member of at least on of the clubs Eagles, Falcons, Tigers and Pythons.  If each club has $100$ students, then the number of students who are members of all the four clubs is at the most .....  If each club has $120$ students, then the number of students who are members of at the most two clubs  is at the least .....  If each club has $140$ students, then the number of students who are members of at least three clubs is at least ....   I came across this question and I tried to solve it using the $4$ circles Venn diagram but it is a very chaotic way to solve this (Too many variables come into picture and it all got messed up) and even I was only able to solve the first question. Any way or method to simplify the solution approach for this question? Any help is appreciated. Thanks!
36294
I am having a few doubt on the interpretation of this problem that I have read on book about interviews questions.  Here the text:  A mythical city contains N=100,000 married couples but no children. Each family wishes to continue the male lane but they do not wish to overpopulate. So, each family has one baby per annum until the arrival of the first boy. Assume that all the children are equally like to be born male and female (and independent). Let $p(n)$ be the percentage of children that are male at the end of the year n. How is this percentage expected to evolve through time?  This is the problem and the solution says that the percentage is expected to remain constant at a level $\frac{1}{2}$.  Thanks in advance; if something is not clear, just ask.
95699
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!
68957
For this question:  A committee of three boys and three girls is to be selected from a class of 14 boys and 17 girls. In how many ways can the committe be selected if:  a.) Ana has to be on the committee.  Ana is a girl so: $$(^{14}C_3)(^{16}C_2)=43680$$  b.) the girls must include either Roberta or Priya, but not both.  I basically needed help in this question (Question B)  Attempts I made so far:  1.) $$(^{14}C_3)*2(^{16}C_2)-(^{15}C_1)=87345$$ 2.) $$(^{14}C_3)*2[(^{16}C_2)-(^{15}C_1)]=76440$$  I need help in what I've been doing wrong in these workings for question B
13089
I know that $$P\left(\bigcup_{i=1}^{n} A_i \right)$$   is the sum of of the probabilities of all the sample points that are contained in at least one of the $A_{i}$'s. This is the probability of sample points belonging to exactly 1 event, exactly 2 events, ...,exactly $n$ events. WLOG this can be written as $$P\left(\bigcup_{i=1}^{n} A_i \right) = P(A_1) + P(A_1 \cap A_2) + \cdots + P(A_1 \cap A_2 \cap \cdots \cap A_n)$$   But the $A_i$'s are arbitrary and we have to account for that. So there are $n$ possibilities for the first probability, $\binom{n}{2}$ possibilities for the second probability, ..., and $1$ possibility for the final probability. So we add and subtract these to prevent overcounting?
93350
What is the probability that in a randomly chosen group of $r$ people at least three people have the same birthday?   $\displaystyle 1- \frac{365\cdot364 \cdots(365-r+1)}{365^r}$   $\displaystyle \frac{365\cdot364         \cdots(365-r+1)}{365^r} +{r\choose 2}\cdot \frac{365\cdot364\cdot363 \cdots         (364-(r-2) +1)}{364^{r-2}}$  $\displaystyle 1- \frac{365\cdot364             \cdots(365-r+1)}{365^r} +{r\choose 2}\cdot \frac{365\cdot364\cdot363 \cdots (364-(r-2) +1)}{364^{r-2}}$  $\displaystyle\frac{365\cdot364 \cdots(365-r+1)}{365^r}                 $     My attempt :  (May be typo in option $(3)$ of question !)  $$P(\text{at least 3 persons have same birthday})$$  $$= 1 - \{P\text{(no one has same birthday) + P(any 2 have same birthday)\}}$$  So, option $(3)$ is true.     Can you explain it, please?   It  before,  but I'm not satisfied by explanation.
11600
I have been teaching statistics for many years at the university level - but this question is beyond my understanding - due to the variables involved.  Several years ago, I was entertaining a relocation to a different area and a different way of life. I was very unsure and skeptical - as the change would have followed my heart - but not the rational practicality of my head.  Then a sequence of events took place - none of which were of my own choosing. I want to know what the statistical probability is of these events occurring by chance alone. They are as follows:  1) The property I had wanted to buy was already under contract, yet came back on the market on the anniversary of my father's death.  2) My immediate offer to purchase was accepted by the sellers (two days later) on the anniversary of my grandmother's death.  3) My loan was approved by my bank on the anniversary of my late grandmother's birthday (7 days later)  4) Our closing date was set by the title company on the anniversary of my late, great-aunt's birthday.  5) The permit for our business (for which we brought the property) was approved and issued on my parent's anniversary.  Can the probability of these dates be statistically determined?
87786
I know that the team has to win 3 of the 4 remaining games. So I thought that from the 4 remaining games, there are a total of $2^4 = 16$ total different outcomes.   Could someone point me towards the right direction to determine the probability of the team winning 3 of the 4 remaining games? Would it be 4C3/16 since the three wins can be chosen amongst the 4 games left?