A $125-$pages document is on the floor randomly spread. If $4$ pages are picked up at random, what is the chance that these are the document's last four pages and in the right order? Page $122, 123, 124$ and $125.$
120466
I have a set of documents numbered 1 to n. If I arrange the documents randomly, what is the chance of selecting the last k pages in the correct order if I select them at random from the documents?   Because order is important, I use the following formula:  $\displaystyle\frac{n!}{(n-k)!}$  From this I calculate how many ways there are to select k pages from n pages in the correct order. I need to have the last k pages. Which I can do in only one way. So is the chance   1/$\displaystyle\frac{n!}{(n-k)!}$?   It seems unlikely. It seems to me spontaneously that the chance should be smaller.
100926
I was looking at this website  and I read the explanation for how to calculate the probability of getting a full house.  To me, the logic basically looked like you figure out the number of possible ranks and multiply by the number of ways to choose the cards from that given rank.  In other words, for a full house $P=$ $$\frac{{13\choose1}{4\choose3}{12\choose1}{4\choose2}}{52\choose5}$$  Following this logic, I tried to calculate the probability of getting two pair.  My (incorrect) logic was that there are 13 possible ranks for the first pair and $4\choose2$ ways to choose two cards from that rank, 12 possible ranks for the second pair and $4\choose2$ ways to choose two cards from that rank, and 11 possible ranks for last card and $4\choose1$ ways to choose a card from that rank.    So I tried $P=$ $$\frac{{13\choose1}{4\choose2}{12\choose1}{4\choose2}{11\choose1}{4\choose1}}{52\choose5}$$  Obviously my solution was incorrect.  I read explanation and the correct answer is $P=$ $$\frac{{13\choose2}{4\choose2}{4\choose2}{11\choose1}{4\choose1}}{52\choose5}$$  I'm still a bit fuzzy on where I went wrong though.  Can anyone help me understand this problem a little better?  Thank you very much for your help.
36179
I am reviewing take aways. I am having trouble  How do I do $342 - 58$?  For the ones column I made the $2$ into a $12$ so I can do $12 - 8 = 4$ but I must take away one tens. So I get $3 - 5$ in tens column but I cant do $3 - 5$. What do I do now do I borrow something else?
85632
Suppose there are $N$ cards, $1,2,\dots,N$.   We start drawing cards (with replacement), until each card has been drawn at least once (we stop when the last card is drawn for the first time).   Let $x$ be the number of cards that have been drawn exactly once.   What is the expected value of $x$?   The answer is supposed to be $1 + \frac{1}{2} +\dots+ \frac{1}{N}$.
39036
"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!
115723
A Professor prepares n letters and puts them in envelopes and closes them. Then realizes he has to write addresses, now he randomly writes addresses in the closed envelopes. What is the probability that he sends exactly k letters with correct addresses?  Attempt: Fix the k correct letters. We can choose them in $ n\choose k$ ways. Total number of combination is n!. Could I say answer is $\frac {n\choose k} {n!}$.  I am just getting started with probability, please post answers considering this.
29781
I have the following problem: A class has 20 students, 16 females, 4 males. Find all possibilities of choosing 6 students so that at least one is male.  I did the following, pick one male from the 4, you have 4 possibilities, then you are left with 19 students, pick 5, so I get 4*${19}\choose{5}$ possibilities.  However, one can notice that if you want no one to be male, then the possibilities are ${16}\choose{6}$, so, one could argue that the solution is instead ${20}\choose{6}$ - ${16}\choose{6}$.  So, which is solution is correct, and why?  $$4* {{19}\choose{5}}$$  or $${{20}\choose{6}} - {{16}\choose{6}}$$
6251
Choose a random number between $0$ and $1$ and record its value. Do this again and add the second number to the first number. Keep doing this until the sum of the numbers exceeds $1$. What's the expected value of the number of random numbers needed to accomplish this?
7371
If I draw three cards at random (without replacement) from a standard 52-card deck, what is the probability that two of the cards will be black and one of them will be red?  Thanks!
96873
Two mathematicians each come into a coffee shop at a random time between 8:00 a.m. and 9:00 a.m. each day. Each orders a cup of coffee then sits at a table, reading a newspaper for 20 minutes before leaving to go to work.  On any day, what is the probability that both mathematicians are at the coffee shop at the same time (that is, their arrival times are within 20 minutes of each other)?
6114
Five bald men wearing hats go to a party, and when they arrived they put their hats in a dark closet. During the party, someone yells, “Fire!” and the men rush to the closet, grab a hat at random, and leave. What is the probability that  (a) exactly four get the right hats?  (b) all five get the wrong hat?  (c) more than one man gets the right hat?
52652
I don’t know how to handle this 3 players case, may anyone give a hand? Thanks!     When playing card game, there are 3 players playing 32 cards, each   receive 10 cards, 2 remind on the table. Overall are 4 aces and 4   kings in the game. How likely is P (E | D)? D = "The table remains 2   aces" E = "I (at least) get one king".   I guess I should have steps as follows.   P(E) = P(1King)+P(2K)+P(3K)+P(4K) P(D) = (4/32) * (4/32) = 1/256 P (E | D) = P(E) + P(D)   Equations may be used:  Counting combinations      The number of combinations of n objects taken r at a time is     nCr = n(n - 1)(n - 2) . . . (n - r + 1)/r! = n! / r!(n - r)!                 Number of favorable outcomes P(I) =     Total number of possible outcome
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?
102177
I'm back again So there's another problem that I can't get to prove  If we take 21 numbers randomly from $1, 2, 3, ..., 40$ then between those $21$ numbers we will be able to find two numbers, of which the smaller one will divide the bigger one  I've been reading james hein "discrete structures, logic and computability" but still can't get to think logically myself. I would be very grateful if I could get some directions/tips/hints or anything, thanks in advance
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.
36415
Question I know this is a duplicate of , i am still posting it because i want to know flaw in my approach.  Candidates were asked to come to an interview with $3$ pens each. Black, blue, green and red were the permitted pen colours that the candidate could bring. The probability that a candidate comes with all $3$ pens having the same colour is  $?$  My Approach Let $E=\text{chossing all three pen of same colour}=\binom{4}{3} \times 1 \times 1 \times 1=4$ i.e select which colour among red,blue,green and black and then there will be only $1$ choice Let $F=\text{possible way of selecting 3 pen}=4 \times 4 \times 4$ As there are $4$ ways to select colour of the pen and we have to choose for $3$ pens. Hence reqd probability$$=\frac{E}{F}$$ $$=\frac{4}{4 \times 4 \times 4}=\frac{1}{16}=0.0625$$ Where am i wrong? Please help!
401
Could someone help me with how the following is calculated:  What is the probability of rolling a die seven times and getting at least one six?  My instinct told me it would be $1/6+\cdots + 1/6$ but this ends up being $7/6$.