Three shooters shoot at the same target, each of them shoots just once. The three of them respectively hit the target with a probability of 60%, 70%, and 80%. What is the probability that the shooters will hit the target  a) at least once  b) at least twice    I have an approach to this but I'm not sure if there's just a formula for this type of thing. For part (a) I was thinking of simply adding the probability of all of the valid scenarios. For instance $$[P(A)*(1-P(B))*(1-P(C))]+[(1-P(A))*P(B)*(1-P(C))]+...$$  and so on until I cover all scenarios in which "at least one" hits. Is there a simpler way? (i assume this would also apply to part b)
123911
Three shooters shoot at the same target, each of them shoots just once. The first one hits the target with a probability of $70\%$, the second one with a probability of $80\%$, and the third one with a probability of $90\%$. What is the probability that the shooters will hit the target  a) at least once?  b) at least twice?  I don't understand well this problem. If each shooter shoots at the target once, how can each of them shoot at it twice?  Do you know how to solve this problem?
67722
An unfair three-sided die. Side $1$ will be rolled with probability $p_1$ , Side $2$ with probability $p_ 2$ and Side $3$ with probability $p_ 3$ . Find a formula for the probability that the $8^{th}$ $3$ will be rolled on the $10^{th}$ roll.  Attempt:  First 9 rolls there are 7 three's.  $$[\binom{9}{7}(p_3)^7(p_1+p_2)^2]\cdot p_3$$ $$=[84\cdot(p_3)^7(p_1+p_2)^2]\cdot p_3$$  any thoughts of this solution?
59451
This puzzle is taken from Mathematical Puzzles: A Connoisseur's Collection [Peter Winkler]. I don't understand the solution. Alice, Bob, and Carol arrange a three-way duel. Alice is a poor shot, hitting her target only 1/3 of the time on average. Bob is better, hitting his target 2/3 of the time. Carol is a sure shot. They take turns shooting, first Alice, then Bob, the Carol, then back to Alice, and so on until one is left. What is Alice's best course of action?  The solution is that Alice is better of missing than hitting Carol or Bob, so she should shoot into the air. Indeed, then Bob will shot Carol, and it can be shown that it gives the greatest probability of survival for Alice. But I wonder if Bob should not voluntary shoot into the air too, so that Carol will do the same, and no one be shot. If this is the case, Alice survival probability is 1. What do you think of it? What is Alice survival probability?
16382
Four cards are face down on a table. You are told that two are red and two are black, and you need to guess which two are red and which two are black. You do this by pointing to the two cards you’re guessing are red (and then implicitly you’re guessing that the other two are black). Assume that all configurations are equally likely, and that you do not have psychic powers. Find the probability that exactly j of your guesses are correct, for j = 0, 1, 2, 3, 4. Hint: Some probabilities are 0. My professor worked out this example in class and I know the answers for j = 1 and j = 3 are zero, j = 0 is 1/6, j = 2 is 2/3, and j = 4 is 1/6 but I do not understand the process or concept behind the question. I do not understand where the numbers are coming from and why only the even js have a probability but not the odd js. The j guesses and pointing is confusing me. Can someone please help explain the question, I would appreciate it.
74362
If $x_1,x_2,x_3$ and $x_4$ are independent identically distributed  random variable with continuous distribution .what is the probability of $$P=p(x_1 &lt; x_2&lt; \max(x_3,x_4))$$ I know the exhaustive number of cases are $p(x_1&lt;x_2&lt;x_3)+p(x_1&lt;x_2&lt;x_4)$ and  every $x$ are iid so both have same probabiltiy so is $2p(x_1&lt;x_2&lt;x_3)$ is right? But  i get the $p(x_1&lt;x_2&lt;x_3) = 1/6$ and then my answer is $1/3$ but the answer is $1/6.$ Can we write $P$ as $p(x_1&lt;x_2&lt;x_3&lt;x_4)+p(x_1&lt;x_2&lt;x_4&lt;x_3)$?
116033
Needs to be solved using conditional probability. I was thinking using the formula for total probability and having the hypothesis $H_i$ - they get to the $i$-th round and ($A$-they meet) $$P(A)=\sum P(A|H_i)P(H_i)$$ I reach a dilemma when trying to find these values, unsure whether to take into account which part of the tournament tree do they meet..  each player has an equal chance of winning.
73966
Here is the problem setup:  Two people play a game of throwing balls at a pin. Sally hits the pin 70% of the times she throws, and Bill hits the pin 40% of the times he throws. The game is won by the person who hits the target first. If Bill throws first each time, find the probability he will win.  With simulation, I get Bill's probability of winning to be around 0.52 (rounding to two decimal places). How to solve this problem analytically?
109689
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.
50680
I'm having trouble with the following problem     There are 4 different colored balls: Red, Green, Black, Blue what is the probability that I will get 2 of the same colored balls by drawing 4 balls with replacement.   I attempted to solve this problem by splitting the question into two parts. First I calculated the probability of drawing 4 balls where I have 2 of one color and 2 of another color. Call this event $A$.  $$P(A) = \frac{4 \times 1 \times 3 \times 1 \times \binom{4}{2}}{ 4^4}= \frac{72}{4^4}$$  Next I calculated the probability where 2 balls are the same color and the other 2 are different. Call this event $B$ $$P(B)= \frac{4\times1\times3\times2\times4P2}{4^4}$$ I understand that if I sum these two probabilities I should get the answer but for some reason the event B was not calculated correctly. Any feedback and help is appreciated.
93145
Three fair six-sided dice are thrown and the dice show three different numbers.  Find the probability that at least one six is obtained.    Im unsure ofwhat type of question this is, I have tried combinations such as 6C1 over 18C3.    But im not sure if this is correct.  Any guidance is much appreciated.    Thank you
77554
I have a problem, I'm learning probability at the moment (I'm a programmer) and starting I have this:  (Source: Minka.) My neighbor has two children. Assuming that the gender of a child is like a coin flip, it is most likely, a priori, that my neighbor has one boy and one girl, with probability 1/2. The other possibilities—two boys or two girls—have probabilities 1/4 and 1/4.  a. Suppose I ask him whether he has any boys, and he says yes. What is the probability that one child is a girl?  b. Suppose instead that I happen to see one of his children run by, and it is a boy. What is the probability that the other child is a girl?    Now my reasoning is: BB = 1/4 = 0.25 BG = 1/4 = 0.25 GB = 1/4 = 0.25 GG = 1/4 = 0.25  So for a., the probability of G I get it just by summing p(B,G) + p(G,B) = 0.5 And for b. p(G|B) = p(G,B)/p(B) = 0.5/0.5 = 1 that is wrong but I'm not getting why.
43091
...2 to 5 questions answered correctly, out of 20 of them? Each question has 5 choices. Probability of getting one right is 1/5. Probability of getting exactly 1 right is ${20 \choose 1} p^1  q^{19}$, with $p=P(\mathrm{right})$ and $q=P(\mathrm{wrong})$  (which I managed to understand and calculate). However how do I calculate for the problem above?
116341
I'd like to ask for help with a task that I've been given.  The task goes like this: In a poker game (52 cards), 5 players are given 5 cards each. What is the probability that both the first and the third player will get a full-house?  It's apparent that the probability of the first player getting a full house is  $\frac{\binom{13}{1} * \binom{4}{3} * \binom{12}{1} * \binom{4}{2}}{\binom{52}{5}}$  Then the second player gets 5 random cards, so $\frac{\binom{13}{1} * \binom{4}{3} * \binom{12}{1} * \binom{4}{2}}{\binom{52}{5}} * \binom{47}{5}$  Now the third player is supposed to get a full house, I'm assuming it should look like this: $\frac{\binom{13}{1} * \binom{4}{3} * \binom{12}{1} * \binom{4}{2}}{\binom{52}{5}} * \binom{47}{5} * \frac{\binom{13}{1} * \binom{4}{3} * \binom{12}{1} * \binom{4}{2}}{\binom{42}{5}}$  The fourth and the fifth player both receive 5 random cards, so the final equation should be:$\frac{\binom{13}{1} * \binom{4}{3} * \binom{12}{1} * \binom{4}{2}}{\binom{52}{5}} * \binom{47}{5} * \frac{\binom{13}{1} * \binom{4}{3} * \binom{12}{1} * \binom{4}{2}}{\binom{42}{5}} * \binom{37}{5}* \binom{32}{5}$  This is how I would solve the task, but I'm not sure that my solution for the third player is correct. Could anyone more experienced than me have a look at my solution?  Also I'm sorry if there is a solution for this type of task on this site, I tried looking for it and I found a lot of poker probability questions, most of them were however focused on only one player and not multiple players, which is my case.
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 $P(A^c\;\cap\;B)+P(A\cap B)$?
2079
After realizing that my knowledge of how probabilities function has waned significantly, I'm wondering what the probability of rolling multiple 6's or higher (6-10) on four 10-sided dice, five 10-sided dice, etc. up to nine 10-sided dice.   As well, what the probability of rolling a 7 or higher, 8 or higher, 9 or higher on the same sets of dice as mentioned above.   Something like; %chance rolling 3 of 6-10 on 4d10, %chance rolling 3 of 7-10 on 4d10, %chance rolling 3 of 8-10 on 4d10, %chance rolling 3 of 9-10 on 4d10, etc.  %chance rolling 3/4/5 of 6-10 on 5d10... etc.  I've tried creating some functions on AnyDice, but I either can't get the function(s) to work, or I fear I may be misunderstanding my results.  How do I make this work?
53633
Question: whats the probability that at least 3 out of a group of n people have the same birthday.  I am confused as to where to start. I know that it would be complementary probability, but even then I dont know exactly.  Would it be;  P(at least 3 have the same bday) = 1 - (|all 3 have different| + (2 have the same|)? Any directions and suggestions?
72596
Given $n$ independent geometric random variables $X_n$, each with probability parameter $p$ (and thus expectation $E\left(X_n\right) = \frac{1}{p}$), what is $$E_n = E\left(\max_{i \in 1 .. n}X_n\right)$$    If we instead look at a continuous-time analogue, e.g. exponential random variables $Y_n$ with rate parameter $\lambda$, this is simple: $$E\left(\max_{i \in 1 .. n}Y_n\right) = \sum_{i=1}^n\frac{1}{i\lambda}$$  (I think this is right... that's the time for the first plus the time for the second plus ... plus the time for the last.)  However, I can't find something similarly nice for the discrete-time case.    What I have done is to construct a Markov chain modelling the number of the $X_n$ that haven't yet "hit". (i.e. at each time interval, perform a binomial trial on the number of $X_n$ remaining to see which "hit", and then move to the number that didn't "hit".) This gives $$E_n = 1 + \sum_{i=0}^n \left(\begin{matrix}n\\i\end{matrix}\right)p^{n-i}(1-p)^iE_i$$ which gives the correct answer, but is a nightmare of recursion to calculate. I'm hoping for something in a shorter form.