probability questions?

a) It is to your advantage to switch. You can improve your odds from 1/3 to 2/3.

b) Depends on whether the chocolates are distributed individually, or in a box, randomly, etc. If they are distributed individually and randomly, the probability is: 0.0002612441329, approximately (edited, again!).

Edit 2: Gianlino was right on this one. Credit where credit is due. However, I used 8C3 instead of 8C5, just to be different :).

c) Depends on how many are sitting at the table. If all of them are sitting at the table, that's a big table, and the probability is: 1/300.

a) This is a classic, non-intuitive problem called the "Monty Hall" problem after the game show host on the old show "Let's Make a Deal" that stumped many a college math professor. By non-intuitive, I mean that many a learned person's intuition told them that the probability had to be 1/2 because with two doors left, it was obvious that it didn't matter if they switched or not, it was one choice, two doors either way. Yet, the probably was actually not 1/2. It was solved publicly by Marilyn vos Savant in a newspaper column. Check out the table on this link: http://en.wikipedia.org/wiki/Monty_Hall_problem

(Edited to provide more explanation)

Door 1-2-3 Possibilities:

Arrangement 1: Car - Goat - Goat

Arrangement 2: Goat - Car - Goat

Arrangement 3: Goat - Goat - Car

Assume you play this game over and over, and the car and goats are arranged randomly each time, but you always pick door 1. No matter, every time you play, the host can always open a door and show you a goat. If it's a row 1 situation, you would have picked the car to begin with, but lose if you switch. If it's a row 2 or 3 situation, you would be wrong to begin with and would win by switching. You lose in one out of three situations by switching, and win in two out of three situations by switching.

b) This can be solved using the binomial theorem by using the fifth-power (1/11) term of the 8th power row of the binomial expansion. Each time a chocolate is handed out, the chance of a particular student getting it is 1/11, and the chance of not getting it is 10/11. In the 8th power row, n = 8. In the fifth power (1/11) term, r = 3 (because the first term is r=0), and n - r = 5. The probability of the student getting five is: (8C3)(1/11)^(5) x (10/11)^(3). The 8C3 is "combinations" (and 8C3 = 8C5).

c) Call the two people seated on either side of the host, A and B.

Case 1: There is a 1/25 chance that A will be seated on the hosts left "and" a 1/24 chance B will be on his right. For "and" probability, you multiply. The chance that both of these events happen is (1/25)(1/24) = 1/600.

Case 2: Or, B could be seated on the left and A on the right. This probability is 1/600 also.

Now, it's Case 1 "or" Case 2. For "or" probability, you add 1/600 + 1/600 = 1/300.

I assumed you would want to see some work, but I had to leave for work, and I was thinking it might be interesting if others tried to answer and got the same or different answers. Anway, I hope this helps.

I disagree on b). Clearly for a student A to receive chocolates at exactly 5 given times is

(1/11)^5 * (10/11)^3.

There are (8 C 5) possible choices for the times; So

(8 C 5) (1/11)^5 * (10/11)^3 is the result;

8*7*6/3*2*1 * 10^3 / 11^8 = 2.61244 10^(-4).

Edit @ Mathematishan: I offered my solution first. So you tell me what's wrong with it and then I'll tell you why yours is way off...=)

Edit 2 So the problem is that your sample space makes no sense; You never pick between distribution. You distribute chocolates; So no matter what you do your sample space will have a size 11^8, coz you have 8 chocolates each with 11 destinations.

When one performs a division you divide the number of positive outcomes by the number of possible outcomes, assuming they all have same probability. But the distributions you chose as your sample space are not equiprobable.... so there is nothing to do.

This is a simple exercise on Bernouilli law with parameter p = 1/11. Think of one student being isolated having a 1/11 at each stage to get the chocolate, and the others having globaly 1 - p =10/11 to get it. How it is dispatched among them is irrelevant.

You coud get the assertion "a pink ball has been drawn" contained in here 2 tactics: Case a million: A pink ball became drawn (a million/6) and the two A and B instructed the fact (2/3 * 4/5). All at the same time it is 8/ninety Case 2: A non-pink ball became drawn (5/6) and the two A and B instructed a lie (a million/3 * a million/5). All at the same time it is 5/ninety So the prospect you have fallen into case a million is: 8/ninety / (8/ninety + 5/ninety) = 8/13

42

a) Would not switch door, it would not improve my chances of getting the car, there are only 2 doors left and it is a 50/50 chance of getting a goat or a car.

However, if the goat behind door number 3 is a cute mini goat, I would settle for it, TY =)

__________________

Hi again Sir, Namasté and all blessings and all due respect to you Sir!

Before you opened door number 3 there was a 1/3 chance that I would get a car and 2/3 chances to get a goat.

You opened door number 3 so now there is 1/2 chance to get the car and 1/2 chance to get a goat, so it does not heighten my chances if I switch choice.

So seriously I would take the mini goat!! I already have a car that I like and I can not drive two cars simultaneously Sir, only Drats can do that, I guess.

Isn't an answer intuitive if no equation can be assigned to it? Why Paul would say the door problem is not intuitive negates the fact that you cannot plug in an equation to decide the outcome.

The correct answer is 42.

42!