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Showing posts from June, 2013

The Elevator Puzzle

Q: A large number of people \(n\) get into an elevator in a skyscraper building which has the same number of floors as \(n\). They have predetermined at what floor they will step out completely at random. Once they step out, the elevator proceeds upwards. What is the probability that nobody steps out at the 3rd floor.

Probability Theory: The Logic of Science

A: In a later post, I'll describe an abstraction that can be used to solve such problems and similar ones. The key is to be able to map statistical problems one runs into to this abstraction (placing \(m\) balls in \(n\) bins).

The probability that a person gets off at floor 3 is \(\frac{1}{n}\) where \(n\) is the number of floors in the building. The probability that the person does not get off at the 3rd floor is thus \(\big(1 - \frac{1}{n}\big)\). The probability that none of the \(n\) people step out at the 3rd floor is given by
$$
 \big(1 - \frac{1}{n}\big)^{n}
$$
In the limiting case, for large \(n\),
$$
\lim_{n\rightarro…

The "Unconscious Statistician"

Q: You scoop up a small, uniformly random, volume \(V\) of sand from a container containing \(V_0\) amount in volume of sand and drop it on a square plate of side \(a\) such that it makes a cone of height \(h\). Next, you drop a small ball vertically somewhere on the plate. What is the average probability that it falls on the cone of sand? (see fig below)

A Book on Statistical Inference

A: First off, the title of this post is not an eye catching phrase. There does exist a "law" in statistics that goes by the name "the law of the unconscious statistician". Googling for that term will yield results! It has to do with the following scenario. Assume you have a random variable \(X\) which follows some distribution \(f(X)\). You want to find the expected value of some other function which takes \(X\) as an argument, call this function \(g\). The law of the unconscious statistician says that you need not find \(g(X)\) explicitly. Instead, you can get to the expected value …

Some Set Theory

Q: A probability puzzle cast from a set theory perspective. You are given, \(P(A\cup B) = 0.8\) and \(P(A\cup \neg B) = 0.85\). What is \(P(A)\)?

Statistical Inference
The Probability Tutoring Book: An Intuitive Course for Engineers and Scientists (And Everone Else!)

A: This is best understood through a Venn diagram. Check out the figure below

In the figure
$$
x = A - B\\
y = A \cap B\\
x + y + z = A \cup B
$$
We are given the following
$$
x + y + z = 0.8\\
x + y + u =  0.85\\
x + y + z + u = 1.0
$$
We need to estimate \(x + y\). If you add the first two equations and subtract the third you get
$$
2(x+ y) + z + u - (x + y) -z -u = 1.65 - 1.0
$$
which in turn yields \(x + y = P(A) = 0.65\)

Some good books to learn the art of probability

Fifty Challenging Problems in Probability with Solutions (Dover Books on Mathematics)

This book is a great compilation that covers quite a bit of puzzles. What I like about these puzzles are that they are all tractable and don't require too much adva…

More Balls from an Urn

Q: An urn contains 3 red balls and 2 yellow balls. A friend draws two balls from it and declares that at least one of those balls is yellow. What is the probability that both balls drawn are yellow?

The Probability Tutoring Book: An Intuitive Course for Engineers and Scientists (And Everone Else!)

A: In order to approach this puzzle, let us draw out tree of possible paths and events. This is shown in the figure below.
The favourable cases where there is at least one yellow ball are the bottom three scenarios. The probability that the second ball is also yellow can be worked out by taking just the 4th case and normalizing it with the bottom three cases. This works out as
$$
P(\text{Both Yellow}) = \frac{\frac{1}{10}}{\frac{1}{10} + \frac{3}{10} + \frac{3}{10}} = \frac{1}{7}
$$


Some good books to learn the art of probability

Fifty Challenging Problems in Probability with Solutions (Dover Books on Mathematics)

This book is a great compilation that covers quite a bit of puzzles. What I like …

Ants on a Bracelet

Q: Four ants are on a round bracelet. They randomly pick a direction and start moving in that direction. All four move at the same speed. What is the probability that at least one pair of ants will collide?

The Probability Tutoring Book: An Intuitive Course for Engineers and Scientists (And Everone Else!)

A: The easiest way to think about this problem is to understand that the ants will not collide only if they all move in the same direction. As there are 4 ants, and each ant can take up any two directions, the total number of combinations of directions that the four ants can take are \(2^{4} = 8\). Of these combinations, only two would result in no collision. These are cases when all four ants move in clockwise or anticlockwise direction. This leads to a total of 8-2 = 6 cases when the ants collide. Thus the sought probability is easily estimated as \(\frac{6}{8} = 75\%\)

Some good books to learn the art of probability

Fifty Challenging Problems in Probability with Solutions (Dover Bo…