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A bridge hand has 13 cards.
Can someone explain these equations?
The probability of holding at least 1 ace...
The probability of holding just 1 ace...
The probability of holding at least 2 aces...
(9C1*5C2)/(14C3) + (9C2*5C1)/(14C3)
I don't understand the 30 to 40 foot thing. How does that affect the answer?
This is from a book. There are 2 graves. You know that at least one of the graves contains the body of a woman. If you randomly select one of the graves and exhume the body, what is the probability that it will be a woman? The book gives the probability as 3/4. I thought that the answer in the book was wrong and should be 2/3. I have come to the conclusion that the book is correct after all and the correct answer is 3/4.
The person in the back sees that the two people in front of him have at least one white hat on. The middle person sees the person in front has a white hat, because if it were black then he would know that his own hat is white, because if his own hat were black, then the person in back would know that his own hat is white. So the person in front knows his hat must be white.
2) 9:00 pm
4) 1992
10 minutes.
Let's say you have five dots in a line and another five dots in a line some distance away.
The total number of ways of drawing five lines, connecting the dots on one side with the dots on the other and each line connecting two dots that aren't connected by any other lines, will give the total number of ways of selecting the two questions, but we can't have the lines going straight across because that would be equivalent to selecting the same question for the same student.
Here's an equation using the inclusion-exclusion principle:
48,120 represents the total number of ways of drawing the five lines with at least one line going straight across.
That isn't what we want. We want the total number of ways of drawing five lines with none going straight across. So...
I remember reading somewhere that if the probability of an event occurring is too small, say 10^-50, then it is basically equal to zero.
In other words, it will never happen given even an infinite amount of time and trials. That seems odd because you would think that
with an infinite amount of trials every event that could happen, will happen, no matter how small the probability. But, let's say I ask you to keep flipping a coin until you get 1,000,000,000,000 heads in a row. It seems that this will never happen no matter how long you try. Maybe this explains part of the St. Petersberg paradox (which is discussed elsewhere on this site). Thoughts anyone?
OK. The 44 permutations for the second question needs to be run through for all possible permutations of the first question which is 120 or 5!. We are counting things twice so we divide by 2.
Hi Relentless,
The chances of at least one match contain other situations where there are more than just pairs of people matching birthdays. Three people sharing the same birthday for example. Maybe that accounts for the discrepancy?
Hi sassytonigirl,
I'm not sure what you did but you put 1.0299682 instead of 0.00010299682. Always remember that the probability of an event occurring is always a number between 0 and 1. Just keep trying. You can do it.
Hi Bobbym,
I have the same answer as you but I used a 20 x 20 Markov chain. The way you got the answer involves some math that is way beyond me.
I'm glad I got the right answer.
I think the general formula, for n people and exactly x pairs is...
So for exactly 3 pairs of people out of 23 having the same birthday the probability is 0.01832728.
That's interesting. I hope I have the correct answer. I will wait for some more people to attempt to answer the problem.
You have two boxes. We will call them box #1 and box #2. Box #1 has 20 wooden disks inside that are numbered 1 to 20. Box #2 is empty.
You have a 20 sided die. You roll the die and whichever number comes up on the die, the disk with that number gets moved to the other box.
So, if it is in box #1 it gets moved to box #2, if it is in box #2 it gets moved to box #1.
Starting with all 20 disks in box #1, on average how many rolls are required for box #2 to contain all of the disks and box #1 is empty?
People buying powerball tickets...
We have a linear Diophantine equation.
9y - 17 = 11x
y-(17/9) = x + (2x/9)
y-x-1 = (8+2x)/9
Because x and y are integers so is (8+2x)/9.
(8+2x)/9=z
8+2x = 9z
4+x=4z+(z/2)
4+x-4z=(z/2)
z=2w
we have
8+2x=18w
so we have
x=9w - 4
y=11w - 3
So we have 2 solutions x=-4 and y=-3 also x=5 and y=8
To get the other number we subtract y from 11.
so we have 83 as one solution and -38 as another solution also -83 is another solution.
In the example where the elements are the same, such as {1,1,1,1}, and you have X=3, the number of ways to get X=3 is simply 4C3. Very interesting problem.
The other problem is this, again 3 people A, B, and C.
A and B working together can prepare a shipment in 8 days.
A and C working together can prepare a shipment in 9 days.
B and C working together can prepare a shipment in 10 days.How long would C take working alone?
The book says 22 days. I am getting 23 and 7/31 days.That is what I am getting also.
Yeah, I don't understand the solution in the book. They have the following equations...
(A + B)/4 = 8
(A + C)/4 = 9
(B + C)/4 = 10
So, A = 14, B = 18, and C =22.
I honestly have no idea what they're doing.
This is back a few pages..
Now supposing 2 die had faces of 2,4,6,8,10,12 and 2 other die had faces of 1,3,5,7,9,11.
All four die are thrown once. What is the probability of them summing to 20 or more?
Is it 545/648?
I have a book which I think has the wrong answers to 2 problems.
In the first problem there are 3 people, we'll call them A, B, and C.
A can make 5 scarves while C makes 2.
B makes 4 scarves while A makes 3.
A's scarf takes 5 times as much cloth as B's scarf.
Three of B's scarves take as much cloth as 5 of C's scarves.
C's scarves are 4 times as warm as B's scarves.
A's scarves are 3 times the C's scarves.
Who's the best overall?
Interestingly, I have 2 books which have this exact same problem, one book is from 1885 and the other book
is from 2013. I believe that the book from 1885 has the correct answer, it claims that the best overall is C.
The book from 2013 claims the answer is A. The book from 2013 makes the mistake (which is pointed out
by the older book) of adding the numbers once the proportions are established, when they should be multiplied instead.
The other problem is this, again 3 people A, B, and C.
A and B working together can prepare a shipment in 8 days.
A and C working together can prepare a shipment in 9 days.
B and C working together can prepare a shipment in 10 days.
How long would C take working alone?
The book says 22 days. I am getting 23 and 7/31 days.
16 x 1.5^2 != 24^2, 16^2 x 1.5^2 = 24^2.
Try moving the 1.5^2 to the right side of the equation and see what you can do.