You are not logged in.
I got 18.75%.
Eduardo, I guess the question is to find the probability that both insects find themselves in one quarter of the circle and in particular 1/8 before and 1/8 after the starting point, during the minute 22 to 23, right?
So let's start by setting the boundaries for each. Where will the spider be after exactly 22 minutes? and after 23? What about the ant?
So, just to clarify:
After moving the 2 apples from A to B, it is ALWAYS unbalanced, unless we are in the bag with the good apples, right?
So, in essence, we can do it in 2 weighings?
Guys, you are geniuses
I could not make it with less than 4...
Thank you!!
Beautiful concept!!
I think 2 weighings are enough.First weigh 5 and 5, leaving out 6.The rest is obvious. I hope it is not homework.
Dear thickhead,
I wish I were a bit younger, even if I had to do some tough homework I am already 36+
But the total number is 17; if we weigh 5 and 5, we leave out 7. Is this what you mean?
Dear phrontister,
All I know is that it can be done with 3.
I have tried several ways: For example, 8+8+1 and we weigh the two sets of 8 together. If they balance, it can be either of the 2:
1. We have the bag with good apples
2. We have the bag with the rotten, and the 1 that was left out is a rotten one, and also that the other 4 are split 2+2.
Second weighting: We pick any of the two sets of 8 (which obviously are identical) and split them into 2 (4+4). If they balance, again it means that each subgroup has 2 rotten (or that we have the bag with the good apples).
But then again we have to make 2 more weightings to deduce... Thus a total of 4
I don't know!
Another idea would be to remove 3 instead of 1, and then split the remaining 14 into 2 groups. If they balance, again it means that the rotten are evenly split into the 2 groups, which means that in the remaining 3, we have an odd number of rotten (either 3 or 1). Then what?
Our local greengrocer had two bags with 17 apples in each. In the first bag, all apples are good, while in the other one, 5 of them are rotten to the core, but seem good to the outside. All the good apples weight the same and all the rotten also weight the same (but have different weight from the good ones). The greengrocer has sold one of the two bags, but without knowing which one! Now he is left with the other bag and wants to find out whether it contains all good apples, or it is the one with the 5 rotten. Using only a balance scale, what is the minimum number of weighings required to determine which bag he has?
If n=1 then we have 1 complete jump in each minute plus some unfinished jumps before and after. Each jump must be exactly 1m, right? Then the MINIMUM is 60m? Not sure.
Of course the maximum is a lot more difficult...
During a game show, the host has placed three contestants (A, B, C) in a closed room. The host places one stamp on each of the contestant’s forehead in such a way that each contestant can only see the other stamps, but not his own. He places a white or black stamp to the forehead of A, a red or green stamp to B and a yellow or blue stamp to C. He then asks each of them to simultaneously write on a piece of paper the color of the stamp they think that they have on their own forehead. The only choices that are permitted are, either a color, or "I don't know". The contestants will win if at least one of them correctly guesses his own color, but with the restriction that no one must not give a wrong answer (for example, if 2 of the 3 contestants reply "I don't know" and the 3rd one guesses correctly, they win the contest).
The host allows the contestants to discuss and coordinate for a few minutes, before the placement of the stamps, in order to decide on a strategy. What is the optimum strategy that will give them the biggest chance to win?
@thickhead: You are a genius!!
The sponsor of a basketball team has offered 209 branded t-shirts to the regional high school players. Each of the t-shirts had a unique number from 2 to 210 printed on its back.
In order to randomly assign the t-shirts to the students, the kids were all given a unique number from 1 to 209 and the shirts would be randomly assigned by carrying out some kind of draw. The maths teacher, however, decided to make it a bit more tricky, so he asked the students to find a way to assign the shirts in such a way, that each student's number is a factor of the t-shirt's number. He then asked them to find how many such ways there are and also to give an example of such an arrangement (we remind that each number has at least two factors, 1 and the number itself).
Thanks bobbym - it doesn't work for me either - it says that they are moving to a new machine!
http://www.mathsisfun.com/combinatorics/combinations-permutations-calculator.html
Is there any permutation calculator under this site? I am unaware of it.
Thanks bobbym, I fully understand and obviously I do not blame or underestimate the usefulness of the calculator of this site!
Unfortunately I do not have any programming skills; otherwise I would, for sure!
Hi guys,
Anyone knows any site to generate permutations / combinations over the 10000000 limit?
The Combinations and Permutations Calculator under this site is a GREAT tool but unfortunately has this limitation (
I don't want just to calculate them (this is easy) but also to GENERATE them.
Many thanks!!
No, the only limitation is that he cannot step back as he walks to the school.
Any clue, guys??
Ooops yes you are right, I did some mistakes in substituting the figures! Thanks
salem_ohio wrote:My approach - but not sure if it is correct...
We must discern the following 4 cases for A and B:
1) They make the same forecast and both are right: Probability P = 4/5*8/9 = 32/45
2) Same forecast and both are wrong: - P = 1/5*1/9 = 1/45
3) Different forecasts and A is right: P = 4/5*1/9 = 4/45
4) Different forecasts and B is right: P = 1/5*8/9 = 8/45In our case, we have A and B give different forecasts. In such cases, A is right with overall probability 4/45, while B is right with overall probability 8/45, that is double (that is, A: 1/3 while B: 2/3).
We are asking for the probability that it rains tomorrow, that is, the probability that B is right. I would therefore say that this is 2/3.
What do you think?You are telling same thing as Relentless but have put the value wrong.
3) must be 4/5*1/10 and
4) 1/5 *9/10
My approach - but not sure if it is correct...
We must discern the following 4 cases for A and B:
1) They make the same forecast and both are right: Probability P = 4/5*8/9 = 32/45
2) Same forecast and both are wrong: - P = 1/5*1/9 = 1/45
3) Different forecasts and A is right: P = 4/5*1/9 = 4/45
4) Different forecasts and B is right: P = 1/5*8/9 = 8/45
In our case, we have A and B give different forecasts. In such cases, A is right with overall probability 4/45, while B is right with overall probability 8/45, that is double (that is, A: 1/3 while B: 2/3).
We are asking for the probability that it rains tomorrow, that is, the probability that B is right. I would therefore say that this is 2/3.
What do you think?
I'll make sure to take my umbrella with me
Solution please??
I think the chance of rain is
In any case, be inclined to expect rain.
There are two weather stations, station A and station B which are independent of each other. On average, the weather forecast accuracy of station A is 80% and that of station B is 90%. Station A predicts that there will be no rain tomorrow, whereas station B predicts rain. What is the probability that it rains tomorrow? We are not asking for the exact probability; we are just asking whether it is more likely to rain or not.
A young boy is going to school from his house in the countryside and has to pass through a small forest. He has a snack with him, which he must eat at the noon break, but he is hungry so he eats it while walking to school.
He realizes, of course, that there will be nothing left to eat at the break!
Fortunately, the forest is full of pears that have fallen off the trees, one every few steps, on his way to school. Some of them are rotten, though, which cannot be discerned from the good ones. Only one pear can fit in his lunch box, so he must take with him exactly one pear. Furthermore, he has with him a small balance scale which he needs for the class of physics. He knows that all good pears have exactly the same weight and also that all rotten pears have different weights from the good ones (can also be different from each other but not necessarily). Knowing also that there are more good pears than rotten, how can he select exactly one good pear without walking back and forth?
For sure, unless you come up with some idea!!
I think the prisoners are executed by now.
Correct!! Thank you!
If it were my money I would lock it up in a cupboard and not invest in these ventures. However if I were to advise somebody for a fee it is o.k.
The investment must be such that the yield from each of these will be same after a year.
let x,y,z be the investment in these avenues. Then
@thikhead:
Sorry, your reply skipped my attention!
YES, perfect solution! Thanks!!
I was struggling to make it with triominoes only!!
It is not explicitly specified but obviously you must spread the amount to all three options; otherwise, if you leave one aside and the other two go bankrupt, then you will have nothing!