Applied Maths Theory....

Rnger

Anybody who is doing applied maths must have done most of the past papers at this stage if they want B+

You have have noticed a trend... 98' was a very difficult paper all together (not many sitters) as was 91'... 7 years seperate these two papers....

Add 7 years to 98' :eek:

*Angel*

A bit far-fetched ain't it? 7 years? I generally wouldn't notice the difficulty of a full paper as I usually do the questions in bulks of topic, I think we're in for a hard hydrostatics this year though as it was simple last year.

I'm glad I don't do Hydrostatics!

*Angel*

Deleted User wrote:

I'm glad I don't do Hydrostatics!

nah you're not really, it's a great topic

Ok I'm not.

I'm sorry.

DsC

i didnt know where to post this but...

If you have a game show , and the contestant is shown three doors.

Behind two doors is a goat. And behind one , a Car.
They Choose One.

The host of the game show then reveals a goat behind one of the other two doors and asks the contestant if they would like to change their mind..

What is the probability of them being right if they change their mind
and what is the probability if they dont?

*Angel*

I don't really like probability, but anyway here is what I think

So the guy picks a door, probability = 1/3 of a car

The host shows that a goat is behind one of the other doors, leaving a probability that the car is behind each door of 1/2

But the guy picked one of two doors so the probability must be 1/2, even if he did change his mind.

Well that's my stupid answer, probably wrong so feel free to diss.

oq4v3ht0u76kf2

Believe it or not... if they change their mind their chance is 2/3 whereas if they keep their original choice it is 1/3. This problem is known as Bertrand's Box Paradox. Paradox because it appears to contravene the elementary laws of probability. It originally appeared in Joseph Bertrand's Calcul des probabilites (1889) however it has since become known as the Monty Hall problem due to it apparently using the same format as his game show on television however Monty Hall denies that his show worked exactly to this format.

Google for either the Monty Hall problem or Bertrand's Box Paradox for a full explanation but my brief one is as follows:

When the contestant chooses a door, there is a probability of 1/3 that they choose the door with the car.

P(Car) = 1/3
Therefore, P(Not Car) = 1 - P(Car) = 2/3

When the host opens a door to reveal a goat P(Not Car) is still 2/3 because when the host reveals a goat it does not affect this probability.

Therefore if the contestant changes their choice, there probability of the door they have chosen containing the car is 2/3. The probability of their original door containing the car is still 1/3.

A          B          C
Car        Goat       Goat

Contestant chooses door C... P = 1/3 of that being car.
Therefore, P of it being behind A or B = 2/3
Host opens door B, so the contestant knows the car is not there...
However, P of the car being behind A or B = 2/3
The contestant KNOWS it is not B therefore P(A) = 2/3

Great paradox!

*Angel*

Bob wrote:

Great paradox!

Yeah, weird...

DsC

Nice one bob , on the money.
Read some Ian Stewart?

oq4v3ht0u76kf2

DsC: Nope... should I?

Zukustious

Bob wrote:

A          B          C
Car        Goat       Goat

Contestant chooses door C... P = 1/3 of that being car.
Therefore, P of it being behind A or B = 2/3
Host opens door B, so the contestant knows the car is not there...
However, P of the car being behind A or B = 2/3
The contestant KNOWS it is not B therefore P(A) = 2/3

Great paradox!

Surely the situation changes once you know what is behind door B. The situation now is a 50/50 chance for the car being behind either a or c. That doesn't seem like a paradox at all.

oq4v3ht0u76kf2

But at that stage you aren't choosing between 2 doors... you are choosing whether to stick with your original choice which has a 1/3 chance of being correct or changing your original choice which you now know has a 2/3 chance of being correct.

Were you to just choose between two doors then yes, obviously, each has a 50% chance of being correct - hence the elaborate foreplay of the paradox.

Zukustious

Bob wrote:

But at that stage you aren't choosing between 2 doors... you are choosing whether to stick with your original choice which has a 1/3 chance of being correct or changing your original choice which you now know has a 2/3 chance of being correct.

Surely that is choosing between door a or c. You've picked c, but have the opportunity to change your mind to a. That is just picking between a and c.

oq4v3ht0u76kf2

Right, let's go again...

You have to choose between 3 doors. Behind one of these doors is a car. Behind the other two there are goats.

You choose door 1.

There is a 1/3 chance that behind this door there is a car.

Now, consider doors 2 and 3 - there is a 2/3 chance that the car is behind one of these doors.

The host opens door 3 and reveals that there is a goat behind it.

Remember, the probability that the car is behind door 2 or 3 is 2/3.

You have just learned that the probability of there being a car behind door 3 is 0. (You have been shown the goat.)

Remember now, the total probability of the car being behind door 2 or 3 is 2/3.

You have seen the probability of it being behind door 3 is 0.

However, P(2) + P(3) = 2/3

P(3) = 0

Therefore, P(2) = 2/3

If you were to stick with your original choice - door 1, you are sticking with your original chance of 1/3 that the car is behind that door.

But, now you know that the probability of it being behind door 2 is 2/3.

So, if you were to change your mind you have doubled your chances of winning to 2/3.

*Angel*

So this all really relates to leaving cert app maths ... :rolleyes:

nobodythere

However, P of the car being behind A or B = 2/3
The contestant KNOWS it is not B therefore P(A) = 2/3

We know it's not behind be, therefore probability of being behind B is 0. Hence B goes out of the equation and probability of being behind A is 1/2

DsC

See.. You are twice as likely to chose a goat originaly.. (2in3 chance)
And by the Host eliminating a goat .. then if you change your answer you have a car

on the other hand , if you chose the car originaly (1in3 chance)
and the host eliminates a goat , you change you have a goat

And yeah bob , you should read , evolving the alien, the magical maze , flatland , the globe .. or not.. some interesting math and philosophy.. in readable form. Makes everything extremly understandable

oq4v3ht0u76kf2

Grasshoppa:

A + B + C = 1

A = 1/3

=> B + C = 1 - 1/3 = 2/3
But, B = 0
=> 0 + C = 2/3
=> C = 2/3

If you still don't believe me, Google for the Monty Hall problem or Bertrand's Box Paradox.

Zulu

*Angel* wrote:

So this all really relates to leaving cert app maths ... :rolleyes:

no it's taken from the novel : a curious incident of the dog in the night-time by mark haddon. well done people

Zukustious

Just got it. Way easier to understand if you use an extreme example like 1000 doors with a car behind one and the rest full of goats. You pick one and then the host opens 998 leaving two left, the car and goat. Odds seem to be way higher that the car is behind the other door.

Pretty deadly.

nobodythere

Bob wrote:

Grasshoppa:

A + B + C = 1

A = 1/3

=> B + C = 1 - 1/3 = 2/3
But, B = 0
=> 0 + C = 2/3
=> C = 2/3

If you still don't believe me, Google for the Monty Hall problem or Bertrand's Box Paradox.

I'm not starting an argument, I just believe that they're looking at it too rigidly with laws of probability. The way I see it is that because the probability of B being the car is 0, it alters the probability of A and C.

A + B + C = 1

A = 1/3

=> B + C = 1 - 1/3 = 2/3
But, B = 0
=> 0 + C = 2/3
=> C = 2/3

Also A + C = 2/3
But C = 2/3 hence A would have to be zero.
You know yourself that the probability of A isn't zero.
Hence we have to remove B from the equation if we know that it's definitely not that. Hence probability of being A or C is 1/2 each...

Once again not starting an argument, I just think you're sticking too rigidly to the logic of it

nobodythere

I resign myself to not understanding and get back to something that WILL come up on my exam :rolleyes:

DsC

Zulu wrote:

no it's taken from the novel : a curious incident of the dog in the night-time by mark haddon. well done people

Whut?

Zukustious

Take the situation with 1000 doors and only one has a car. You have to pick one. The odds of you getting the car are 1/1000. 998 doors are opened to reveal goats and you have the freedom to choose your original door, or a different one. Playing this out 1000 times, you'll only get the car in the original door once(most likely), so odds are the car is in the other door that you're left with.

oq4v3ht0u76kf2

It features in the book, though I first heard of the paradox in a biography of Bertrand. The book does provide a reasonable explanation of the problem though.

DsC

See the thing to understand is that in probability.. we have conditionals
and because the conditional (the host revealing a goat) happened after the Fact (you picking the door) its different. You dont choose a door from 1of2 , then it would be fifty fifty.. you choose from 1 of 3 then get more information , then make a new choice.

Anywho , I reckon if you get a goat your a winner anyway