TheAspiringDoc's Math Thread

[keltingmeith]

You need to use curly brackets to get multiple things in exponents, btw. You should try quoting people's posts to see how they've acheived writing certain things, and then use their method of typesetting to do it yourself.

Also, a couple of things you might be interested in:

1. The Bernoulli and Binomial distributions. These can be used to model certain problems similar to the one in your first question.

2. Modular arithmetic. Believe it or not, you actually did this in your second question - it is an actual mathematical thing used in algebra and number theory, you might enjoy a read about it.

[TheAspiringDoc]

Wow, the more you look into this problem, the harder it is.

A hundred people board a plane (tha has 100 seats) with assigned seating. For some reason the first person on board takes a random seat. The second person looks around and if their seat is taken, takes a random seat and otherwise sits in their assigned seat. The third person does the same and this goes on for all remaining passengers. What is the probability that the 100th person sits in their assigned seat?

Help?
Thunks

[keltingmeith]

Wow, the more you look into this problem, the harder it is.

A hundred people board a plane (tha has 100 seats) with assigned seating. For some reason the first person on board takes a random seat. The second person looks around and if their seat is taken, takes a random seat and otherwise sits in their assigned seat. The third person does the same and this goes on for all remaining passengers. What is the probability that the 100th person sits in their assigned seat?

Help?
Thunks

Try doing the problem for 2 or 3 people, first. THEN try putting it up to 100.

The problem itself isn't terribly difficult, but the moment you use a big number, is the moment it becomes incredibly daunting.

[kinslayer]

If all seat configurations are equally probable then the probability should be 1/100.

Another way of looking at it: the probability is equal to the probability that the first 99 people boarding the plane choose a seat that is NOT the 100th person's seat.

EDITED:

If there are n seats on the plane (n>1), the probability that the (n-1)'th person will take a seat that is not theirs is 1/2, the probability that the (n-2)'th person chooses a seat that is not theirs 2/3, etc. The probability that each of the n-1 people choose a seat that is not their own is equal to



So the probability that 99 people do not pick the 100th person's seat (the case n = 100) is equal to 1/100.

[TheAspiringDoc]

If all seat configurations are equally probable then the probability should be 1/100.

Another way of looking at it: the probability is equal to the probability that the first 99 people boarding the plane choose a seat that is NOT the 100th person's seat.

If there are n seats on the plane (n>1), the probability that the (n-1)'th person will take a seat that is not theirs is



So the probability that the 99th person takes takes a seat that is not theirs is 1/100, which is the probability you are after.

I'm pretty sure that answer is incorrect.
Think of it in terms of how the last person could not get their seat.
The probability that the wild person at the start chooses the final person's seat straight up is 1/100. Then we must also add the probabilities that the wild person at the start chooses some other person's seat, who in turn takes the final persons seat.
These two alone add up to more than 1/100. And there are plenty of others too.

[lzxnl]

I'm pretty sure that answer is incorrect.
Think of it in terms of how the last person could not get their seat.
The probability that the wild person at the start chooses the final person's seat straight up is 1/100. Then we must also add the probabilities that the wild person at the start chooses some other person's seat, who in turn takes the final persons seat.
These two alone add up to more than 1/100. And there are plenty of others too.

You misinterpreted what kinslayer wrote. In your defence he didn't spell everything out
He worked out the chance that the 100th person wouldn't sit in their assigned seat. So the answer is 0.99

[TheAspiringDoc]

You misinterpreted what kinslayer wrote. In your defence he didn't spell everything out
He worked out the chance that the 100th person wouldn't sit in their assigned seat. So the answer is 0.99

Wait, you're sating that the chance that the 100th person sits in their seat is 0.99?
That doesn't seem right..

[keltingmeith]

I haven't actually gone through this yet because I have no free time, but I do agree that the number seems a bit high. Having said that, though, this doesn't mean that the answer is wrong - consider the birthday paradox.

Basically, it is very common to come up with mathematical results that don't seem correct, which is why we have a bunch of results collectively known as "paradox"es. It basically means that just because your intuition tells you one thing, don't expect it to be true unless the maths tells you it is. In converse, just because something seems wrong, if the mathematics is correct, that means it must be right.

[kinslayer]

I'm pretty sure that answer is incorrect.
Think of it in terms of how the last person could not get their seat.
The probability that the wild person at the start chooses the final person's seat straight up is 1/100. Then we must also add the probabilities that the wild person at the start chooses some other person's seat, who in turn takes the final persons seat.
These two alone add up to more than 1/100. And there are plenty of others too.

You're right, I misinterpreted the question. My solution assumes everyone chooses a set randomly, when in fact they only choose randomly if their seat is already taken. The probability should be much higher than 1/100. It wouldn't be as high as 0.99, but perhaps not far off. I will revisit when I have more time.

[lzxnl]

Yeah. I read through my interpretation again and it's flawed as well.

Repeating question for my own convenience.
A hundred people board a plane (that has 100 seats) with assigned seating. For some reason the first person on board takes a random seat. The second person looks around and if their seat is taken, takes a random seat and otherwise sits in their assigned seat. The third person does the same and this goes on for all remaining passengers. What is the probability that the 100th person sits in their assigned seat?

Let's break this problem down a bit. Let P(n) be the chance the nth passenger gets his seat.
With n seats, there are a few cases.
The first person can, with 1/n probability, choose his own seat. This then forces all of the other passengers to get their own seat.
The first person can, with 1/n probability, choose the nth person's seat. This leads to a null result.
They also can, with 1/n probability, choose any other person's seat. This results in the same dilemma with fewer passengers. If the person chooses the seat x from the end, then all of the people from 2 to n-x will have to choose their own seat. The (n-x+1)th person then has to pick a random seat, beginning the dilemma again.
So the chance that the nth passenger will get his own seat is the probability the first event occurs * 1 (as it's then guaranteed that person n gets his seat) + probability third event occurs * (sum of all the different dilemma probabilities)
In other words, P(n) = 1/n + (sum (P(i))/n) where the sum is taken from i = 2 to n
P(n-1) = 1/(n-1) + 1/(n-1)* sum(P(i))  where the sum is taken from i=2 to n-1
n*P(n) = 1 + sum(P(i)), 2<i<n
(n-1)*P(n-1) = 1 + sum(P(i)), 2<i<n-1
n*P(n) - (n-1)*P(n-1) = P(n)
(n-1)*P(n) = (n-1)*P(n-1)
P(n) = P(n-1)
By induction, as P(2) = 1/2, P(n) = 1/2 for all n

[keltingmeith]

In other words, P(n) = 1/n + (sum (P(i))/n) where the sum is taken from i = 2 to n

I'm really not sure how you've managed to arrive at this conclusion, care to walk me through it some more?

[lzxnl]

P(n) = chance that nth person gets his own seat

Person n gets his seat if the first person to choose gets any seat that's not n. The first person has chance 1/n to pick seat 1, which necessarily means person 2 gets seat 2 and so forth, so this option contributes 1/n * 1 = 1/n
Now person 1 has a 1/n chance of choosing any one of the seats from 2 to n-1. Let the person choose seat i from the end (end being seat 1). Then, all of the people from person 2 to person n-i must choose their own seat. For the i people from n-i+1 to n, they have a similar selection dilemma as the original situation, so the probability in the latter case of the nth person choosing his own seat is P(i). Each choice contributes 1/n * P(i). Hence the summation.

Although I realised that the sum should be taken from 2 to n-1 here. It doesn't actually make much of a difference though:

P(n) = 1/n + (sum (P(i))/n) where the sum is taken from i = 2 to n-1
P(n-1) = 1/(n-1) + 1/(n-1)* sum(P(i))  where the sum is taken from i=2 to n-2
n*P(n) = 1 + sum(P(i)), 2<i<n-1
(n-1)*P(n-1) = 1 + sum(P(i)), 2<i<n-2
n*P(n) - (n-1)*P(n-1) = P(n-1)
(n)*P(n) = (n)*P(n-1)
P(n) = P(n-1)
By induction, as P(2) = 1/2, P(n) = 1/2 for all n
As required anyway

[TheAspiringDoc]

The expression 15^80 x 28^60 x 55^70 gives a number that ends with a string of zeros.

How many consecutive zeros are in that consecutive string?

Hand-held calculator allowed.

Thanks

[lzxnl]

The expression 15^80 x 28^60 x 55^70 gives a number that ends with a string of zeros.

How many consecutive zeros are in that consecutive string?

Hand-held calculator allowed.

Thanks

Prime-factorise everything.
15 = 3*5
28 = 2^2 * 7
55 = 5*11
Taking the necessary powers gives (3^80*5^80) * (2^120 * 7^60) * (5^70*11^70) = 3^80 * 5^150 * 2^120 * 7^60 * 11^70
The logic is that zeros only come from powers of 10 in the prime factorisation.
I can clearly take out a maximum of 10^120 here so there are 120 zeroes.

[MathsGuru]

Ok, thought my math's posts were a bit out of place in the mainstream question threads so I thought I'd make my own
I'm trying to get an intuitive grasp of the derivatives and antiderivatives of trigonometric functions.. Any idea's how?
Also, Find the length of the largest square that will fit inside a regular hexagon of edge length 30cm?

Another way to see why the derivative of sin(x) is cos(x), and why that of cos(x) is -sin(x), is to look at their power series, if you're familiar with what they are (sort of "infinite degree polynomials", roughly speaking).

This is more algebraic than intuitive though.