Say there are 12 different kinds of toys available and you don't know which one you'll get as a free gift when you buy a Happy Meal.  How many Happy Meals do you have to order before you can collect the whole set? Assuming the toys are allocated randomly and the number produced is the same for each toy.
121774
Assume last year I commuted to work in a taxi, and suppose if there are $n$ taxis in the fleet I used. If I took one of these taxis every trip at random and with replacement, then what is the number of trips I need to make in order to use every taxi in the fleet?
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.
37252
A student prepares for an exam by studying a list of 12 problems. She can solve 9 of them. For the exam, the instructor selects 7 problems at random from the 12 on the list given to the students. What is the probability that the student can solve 5 of the 7 problems on the exam?  I am really confused by this problem. My first initial thought was to do:  (5 chose 7)(7 chose 0)/ (12 chose 7)  I am not 100% sure about this though.
45915
How many ways are there to distribute 5 distinguishable balls between 3 kids such that each of them gets at least one ball?   My approach is $ \binom{5}{3} 3! $ + $ \binom{2}{2} \binom{3}{2}2!$ which is equal to $66$?
7424
In how many ways can we choose k items from n distinct items put in a circle.   Then further extension of the question, if I want to find out the number of ways to choose k items from n distinct items in a circle such that no 2 are together.  Thanks
76514
folks,  i am new to this forum and not a math expert. so please bear with me if am asking silly questions.   The question is "probability of getting 50 heads from tossing a coin 100 times".  So the answer for this is, I guess, ${100 \choose 50} (2 ^{-100})$.  So all am trying to get is easier way to calculate ${100 \choose 50}$, or another approach to the parent problem only.  Thanks all, appreciate that.  raj
100343
The questions are: a) There are $5$ comic books, $3$ cooking books, $2$ grammar books. In how many ways can these books be arranged on a shelf if no two of the $3$ cooking books are together?   b) There are $5$ comic books, $3$ cooking books, $2$ grammar books. In how many ways can these books be arranged on a shelf if one of the grammar books is on the left of all cooking books, and the other grammar book is on the right of all cooking books.  For b) after calculation I got $5!\times2!\times10\times9\times8\times7\times6$. Did I solve it correctly?  For a) I'm kind of confused... Any help will be very much appreciated! Thanks
21919
I have been looking at the birthday problem (http://en.wikipedia.org/wiki/Birthday_problem) and I am trying to figure out what the probability of 3 people sharing a birthday in a room of 30 people is. (Instead of 2).  I thought I understood the problem but I guess not since I have no idea how to do it with 3.
114319
This feature has existed for a while (I'm not sure since when exactly), but I'm mainly interested in the Gen VI implementation (hence the tag).  Usually, at the start of a game, you have some number of Boxes available. In B/W, those were 8, in X/Y I think it was 15. Under certain conditions though, you receive additional Boxes that just appear at the end of your current Boxes list. Which brings me to Part 1 of my question:     1) What are the exact conditions for additional Boxes to appear and do you always get 8 additional Boxes? (Using Pokemon Bank reveals that in X/Y, the maximum number of Boxes is 31, which I have experienced to be unlocked in 2 steps of 8 Boxes each.)   Since I'm a person who really likes to order and organize stuff, my Boxes have names and are used to categorize my Pokemon. This results in some Boxes ending up empty for a while most of the time. So here's to     2) Do additional Boxes you receive stay forever or do they disappear if the conditions for them are no longer met?      3) If they disappear, is there a way (such as renaming or putting any Pokemon in the last Box) to make them stay no matter what, even if you released all your other Pokemon?
22135
I received a word problem that goes like this.  A local kindergarten is thinking of making posters that show all the different ways of adding two or more integers from 1 to 9 to get a sum of 10.  If there is enough space on each poster for up to 50 possible solutions, how many posters will the school need to make?  (Note:  sums that contain the same number but in a different order are considered to be different; for example,  1 + 9  and  9 + 1  are two different solutions.)  What is the answer to this problem, and more importantly, how do I solve it?
60096
15 coupons are numbered 1,2,3...15. 7 coupons are selected at random one at a time with replacement. Find probability that 9 is the largest number appearing on the coupon. I am having a problem finding $N(a)$ for this problem.  My working:  $N(s)= 15^7$ (since you have 15 choices for each time you pick up a coupon)  $N(a)= 9^7$  (since you have 9 permissible choices for each time you pickup a  coupon)  But according to my textbook $N(a)= 9^7 -8^7$  but no further explanation is given.  So i would really appreciate it if someone could point out where i am going wrong. Thanks in advance :)
60901
You have one penny, one nickel, one dime, and one quarter. How many different amounts of money can you make using one or more of these coins? Please help me! I'm having trouble! I'm having trouble I know the obvious ones like $5,10,25,1,15,26,36,16,\ldots$ Help? And I need to know the coin combinations too. I found $15$ different ones! Anyone who can find more??
78459
So the question is how many ways are there to put N unlabelled balls in M labeled buckets.  One way (I think) I figured it out is $M^N$ ways to put each ball, divided by $N!$ permutations of ball ordering we don't care about:  $$M^N \over N!$$  First, is this correct?  Second, is there another way to approach it without using division?  I need to calculate it in a non-prime modulo space so division isn't available.  I really need a recurrence.
26565
4 cards are drawn from a pack without replacement. What is the probability of getting all 4 from different suits?  Here's how I tried to solve:  For the first draw, we have 52 cards, and we have to pick one suit. So, probability for this is $\frac{13}{52}$.  For the second draw, only 51 cards are left. The second suit has to be selected, so there are 13 cards from that suit. The probability is $\frac{13}{51}$.  Similarly, the third and fourth draw have probabilities $\frac{13}{50}$ and $\frac{13}{49}$ respectively.  Since the draws are independent, the total probability becomes $$\frac{13}{52} \times \frac{13}{51} \times \frac{13}{50} \times \frac{13}{49}$$  But my book says the answer is $\frac{{13\choose 1} \times {13 \choose 1} \times {13\choose1} \times {13\choose1}}{52 \choose 4}$.  My answer differs by a factor of $4!$. What did I do wrong?
76744
Assume you have 52 cards in a deck, and you have to draw 10 cards. Assuming you have to put back the card after each draw, what is the probability that you will not draw any repeated cards?  Here's what I was thinking...   Draw 1: $\frac{52}{52}$ since first card is allowed to be any card. Draw 2: $\frac{51}{52}$ since you don't want to draw the first card, you're now allowed to draw 51 of the remaining cards.  ... Draw 9: $\frac{44}{52}$ since you don't want to draw the last 8 cards, you're now allowed to draw 44 of the remaining cards. Draw 10: $\frac{43}{52}$ since you don't want to draw the last 9 cards, you're now allowed to draw 43 of the remaining cards.   So $\frac{52}{52} \times \frac{51}{52} \times ... \times \frac{44}{52} \times \frac{43}{52} = 0.397$ or $39.7\%$ chance of not getting a repeat.  Just wanted to check if this is the correct approach to the problem or not.
21485
I'm trying to solve the well known Coupon Collector's Problem by explicitly finding the probability distribution (so far all the methods I read involve using some sort of trick). However, I'm not having much luck getting anywhere as combinatorics is not something I'm particularly good at.   The Coupon Collector's Problem is stated as:  There are $m$ different kinds of coupons to be collected from boxes. Assuming each type of coupon is equally likely to be found per box, what's the expected amount of boxes one has to buy to collect all types of coupons?  What I'm attempting:  Let $N$ be the random variable associated with the number of boxes one has to buy to find all coupons. Then $P(N=n)=\frac{|A_n|}{|\Omega _n|}$, where $A_n$ is the set of all outcomes such that all types of coupons are observed in $n$ buys, and $\Omega _n$ is the set of all the possible outcomes in $n$ buys. I think $|\Omega _n| = m^n$, but I'm not even sure about that anymore, as all my attempts so far led to garbage probabilities that either diverged or didn't sum up to 1.
69633
Let's say you tell me to produce randomly a number from 1-100, and I choose the number 47. Can it be said that there is a specific reason I chose the number 47, and that it is not completely random? By random I mean that there is absolutely no reason this number in particular was chosen.    It seems odd to me, since it seems like since I have applied conscious effort to produce or choose a number within that range, or even without range, that it can't be truly random, but there might be a reason why I chose that number amongst the many others I could have chosen.  I guess the general question, which I'm interested to know if any philosophers in the past have discussed, is whether our thoughts are necessarily based on prior thoughts, or if thoughts can be truly spontaneous and random?