In MIT Opencourware Probabilistic Systems Analysis and Applied Probability course, there's a recitation problem for Lecture 4:     Quotestion: Imagine that 8 rooks are randomly placed on a chessboard. Find the   probability that all the rooks will be safe from one another, i.e.   that there is no row or column with more than one rook.   And the solution is like that:      Solution: We will count the number of favorable positions in which we can safely   place 8 rooks, and then divide this by the total number of positions   for 8 rooks on a 8 × 8 chessboard. First we count the number of   favorable positions for the rooks. We will place the rooks one by one.   For the first rook, there are no constraints, so we have 64 choices.   Placing this rook, however, eliminates one row and one column. Thus   for our second rook, we can imagine that the illegal column and row   have been removed, thus leaving us with a 7 × 7 chessboard, and thus   with 49 choices. Similarly, for the third rook we have 36 choices, for   the fourth 25, etc... There are 64 · 63 · · · 57 total ways we can   place 8 rooks without any restrictions, and therefore the probability   we are after is: 64 · 49 · 36 · 25 · 16 · 9 · 4 / (64!/56!)   I understand the part that placing the rooks that all will be safe but I could not understand the part that we find the sample space(total number of positions for 8 rooks on an 8 × 8 chessboard). Why don't we just say that there are C(64,8) ways to choose the cells that we will place the 8 rooks? For example, placing a1 first and e7 second does not matter for us, it could be e7 first and a1 second. It means the ordering is not important for us. So why does the solution counts step by step?
121401
Two rooks are placed on an 8 by 8 chessboard such that they're in different squares. What is the probability they can't attack each other? Rooks can attack each other if and only if they're in the same row or column. I know that there are questions like this but they give the general numbers not the probability I am not sure how to start the questions and I have looked at this questions for hours. Any help would be welcome.
23898
Edit: Sorry for not providing more on this question the first time. I tried to solve it. When choosing 13 cards from a deck of 52 cards, what is the probability of choosing precisely a card with a number 6,7,8 of the same suit? For example a 6,7,8 of heart  or a 6,7,8 of diamonds, or a 6,7,8 of spades, or a 6,7,8 of clubs. All three cards must be the numbers of 6,7,8 and only of a single suit. These must be among the hand of  13 cards. My first thoughts on this question that I only have 4 possibilities of this happening for the entire deck of 52 cards. So I thought $(_4 C_1)(_{13}C _{3})$ will account for the possible number of decks that will include cards of 6,7,8 of all the same suit out of 13 cards in the hand. And $(_{52} C_{13})$ gives me all the possible combinations of 13 cards from 52 cards.  So this probability that I got would be: $\frac{(_4 C_1)(_{13}C _{3})}{(_{52} C_{13})}\approx 5*10^{-7}$. However I feel something is off.
5024
Problem: Find the number of twelve-card hands containing four cards in each of three suits.   My solution:   pick a suit : 4 ways. choose 4 cards from the 13 in that suit: ${^{13}}{C}_{4}$ ways. pick another suit: 3 ways. choose 4 cards from the 13 in the new suit: ${^{13}}{C}_{4}$ ways. pick the final suit: 2 ways. choose 4 cards from the 13 in the final suit: ${^{13}}{C}_{4}$ ways.   Hence, total number of ways is $4\cdot{^{13}}{C}_{4}\cdot3\cdot{^{13}}{C}_{4}\cdot2\cdot{^{13}}{C}_{4} = {^{4}}{P}_{3}\cdot\left({^{13}}{C}_{4}\right)^3 $ total hands.   This is the incorrect answer. Why?
32202
A woman has $n$ keys, one of which will open a door. a)If she tries the keys at random, discarding those that do not work, what is the probability that she will open the door on her $k^{\mathrm{th}}$ try?  Attempt: On her first try, she will have the correct key with probability $\frac1n$. If this does not work, she will throw it away and on her second attempt, she will have the correct key with probability $\frac1{(n-1)}$. So on her $k^{\mathrm{th}}$ try, the probability is $\frac1{(n-(k-1))}$ This does not agree with my solutions.   b)The same as above but this time she does not discard the keys if they do not work.  Attempt: We want the probability on her $k^{\mathrm{th}}$ try. So we want to consider the probability that she must fail on her $k-1$ attempts. Since she keeps all her keys, the correct one is chosen with probability $\frac1n$ for each trial. So the desired probability is $(1-\frac{1}{n})^{k-1} (\frac1n)^k$. Again, does not agree with solutions.  I can't really see any mistake in my logic. Can anyone offer any advice? Many thanks
88569
I'm trying to find the probability of getting 8 trials in a row correct in a block of 25 trials, you have 8 total blocks (of 25 trials) to get 8 trials correct in a row. The probability of getting any trial correct based on guessing is 1/3, after getting 8 in a row correct the blocks will end (so getting more than 8 in a row correct is technically not possible). How would I go about finding the probability of this occurring? I've been thinking along the lines of using (1/3)^8 as the probability of getting 8 in a row correct, there are 17 possible chances to get 8 in a row in a block of 25 trials, if I multiply 17 possibilities * 8 blocks I get 136, would 1-(1-(1/3)^8)^136 give me the likelihood of getting 8 in a row correct in this situation or am I missing something fundamental here?
39087
I am struggling with this probability/combinatorics problem.  the problem states: "11 people are seated around a round table. At some point, everyone stands up and sits again at a randomly chosen seat. Show that the expected value of the number of pairs which swapped places (A sat at B's place and B sat at A's place) is 1/2."  I struggle with calculating the number of orderings in which pairs switch places. I know we have 10! possible permutations for 11 people sitting in a circle, and there are $11\choose2$ possible pairs that could switch places. But I'm not sure how to count for example the number of permutation in which only one couple switches places.
29294
How would you do the following problem:     Suppose that $n$ balls are placed at random into $n$ boxes. Find the probability that there is exactly one empty box.   I mentally pictured $n$ boxes being in front of me and $n$ balls which I'm throwing (hence randomness) in the boxes, each ball has an equal chance of landing in any of the boxes. That led me to this: $$\frac{n(n-1)\frac{n!}{2}}{n^n}\tag{1}$$ Which is a good answer yet to a different problem, and the answer to the above problem turned out to be $$\frac{n(n-1)}{\binom{2n-1}{n}}\tag{2}$$  I can reverse engineer the correct answer $(2)$ and think about it in terms of stars and bars but it's unclear what mental picture do I need to have in mind after reading the word problem?  And as a side question: do you think the phrasing of the cited problem is unambiguous? How would you state the problem which has $(1)$ as its solution?
93686
I am taking a probability course and we have been of course learning the classic problems of men with hats or variations of that in some other form. The problems being of the type if $n$ men have $n$ hats and mix them all together, whats the probability $0, 1, 2,$ or $k$ men choose their hat.   But from what I can gather is that they are all problems of the same type, for any given random permutation of $k$ objects how many fixed points are there. My professor taught us to solve this using conditional probabilities (conditioning on whether the first person matched and then generating a system of equations) and using the inclusion-exclusion principle. But is there any other way to reason this?
26788
There are 50 numbered objects from 1 to 50. We extract 3 of these objects one after the other without reinsertion. Calculate the probability P(a&lt;b&lt;c)  Given 3 objects, they can be ordered in $3!$ different ways. Of all these ways only 1 satisfies a&lt;b&lt;c so the probability should be:  $P(a&lt;b&lt;c) = \frac{1}{3!} = \frac{1}{6}$  Is it solved in the right way?
109117
If $S = \{ A_{m \times n} : m &gt; n, A_{ij} \in \{0,1\} \}, $ is the set of all $m \times n$ binary matrices, and I choose a random matrix $r \in S$, what would be the probability that $r$ would have full column rank (each of the n columns would be linearly independent)?  Several papers simply claim that this probability is very high, but I'm curious as to what the exact probability would be, and how to compute it.
714
What is wrong with my method of find the probability of a 2 pair?  For the first card we have 52 choices, one this is picked( call it A) we have 3 choices(different suite than A but this same value) for the second card, ( lets call this card B). Now we have AB, to we have to throw away 2 more cards so we don't have 3 of a kind or a four of a kind. Therefore one our third card we have 48 choices and on our fourth card we have 3 choices again similar to above.  On our last card we only have 44 choices left. So the probablity of getting a 2 pair should be (52*3*48*3*44)/(52 choose 5).  But this isn't correct.     Note: I edited this to "2 pair"; the original version was wrong.
38535
Say there are $p_{1}$ red balls and $p_{2}$ green balls. We put all the balls in a circle with $p_{1}+p_{2}$ places in total. It is forbidden that a ball (red or green) is placed between two red balls.    Show that this is impossible if $p_{1}&gt;p_{2}$.   Further on, say $p_{1}=p_{2}=p\geq10$. For which values of $p$ the ordination is possible?  Can somebody help me? Thanx in advance!
48908
Consider the squares of an $8 \times 8$ chessboard filled with the numbers $1,2,3,4 \ldots ,64$ in sequential order.   If we choose $8$ squares with the property that there is exactly $1$ from each row and exactly $1$ from each column, and add up the numbers in the chosen squares, show that the sum obtained is always $260$.  Please provide full solution and explanation.
30867
I know that the formula for counting the number of ways in which $n$ indistinguishable balls can be distributed into $k$ distinguishable boxes is $$\binom{n + k -1}{n}$$  but I am having a hard time understanding why this formula counts that. I mean, suppose we have $4$ boxes and $3$ balls, then the problem is equivalent to count the permutations of  5 vertical lines with 3 circles except that two lines have to be fixed (the first and last lines). I would appreciate if someone could help me to relate and translate this way of thinking the problem to the formula with this. Thanks in advance.
19771
I'm confused about why we should change door in the Monty Hall Problem, when thinking from a different perspective  gives me equal probability.   Think about this first: if we have two doors, and one car behind one of them, then we have a 50/50 chance of choosing the right door.   Back to Monty Hall: after we pick a door, one door is opened and shows a goat, and the other door remains closed. Let's call the door we picked A and the  other closed door B. Now since 1 door has already been opened, our knowledge has changed such that the car can only be behind A or B. Therefore, the problem is equivalent to: given two closed doors (A and B) and one car, which door should be chosen (we know it's a 50/50 thing)? Then, not switching door = choosing A, and switching door = choosing B. Therefore, it seems that switching should be equally likely, instead of more likely.   Another way to think: no matter which door we choose from the three, we know BEFOREHAND that we can definitely open a door with a goat in the remaining two. Therefore, showing an open door with a goat reveals nothing new about which door has the car.   What's wrong with this thinking process? (Note that I know the argument why switching gives advantage, and I know experiments have been done to prove that. My question is why the above thinking, which seems legit, is actually wrong.) Thanks.
67864
Completely shuffle and then lay out a full deck of 52 cards in a single row (i.e. single row 52 columns). Do the same thing with a second deck of playing cards and place this deck in the second row(also has 52 columns). What is the expected total number of matches in all the columns?   A match means that the top card and the bottom card in the column are the same (same rank and suit).  I was trying to solve it by brute force but seems like it's not that easy to me.
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I was trying to answer      Find the probability of getting a full house from a $52$ card deck. That is, find the probability of picking a pair of cards with the same rank (face value), and a triple with equal rank (different from the rank of the pair of course).   My idea was this:  Take one rank. The number of ways you can get a pair from a rank is $C(4, 2)$.   Take another rank. The number of ways you can get a triple from that rank is $C(4, 3)$.   Therefore, the number of ways you can get a full house from just $2$ ranks is $C(4, 2) \cdot C(4, 3)$. But there are $13$ different ranks, and the number of ways you can select $2$ ranks out of $13$ is $C(13, 2)$. For each of these pairs of ranks, you can have $C(4, 2) \cdot C(4, 3)$ full houses. Therefore the number of possible full houses is $C(4, 2) \cdot C(4, 3) \cdot C(13, 2)$, and the probability of selecting a full house is  $\dfrac {C(4, 2) \cdot C(4, 3) \cdot C(13, 2)}{C(52, 5)} \approx 0.000720288$  However, this is way lower than the actual answer of $.00144$. Where did I go wrong here?