Probability Puzzles

Q: Four teams are selling selling lemonade to customers walking out of a store. A total of 10 customers walk out of the store and randomly pick a team to buy lemonade from. What is the probability that all teams get at least one customer? Probability Theory: The Logic of Science A: For simplicity, assume that the customers are walking out in a line and pick a team to buy lemonade from as shown in figure below. The probability that a given team would get at least one customer is one minus the probability they get none. For a given customer and a given team, there is a \(\frac{4  - 1}{4} = \frac{3}{4}\) probability that they will not connect. The probability that none of the 10 customers would pick a given team is \(\big(\frac{3}{4})^{10}\). Moving on, the probability that exactly one team would be without customers is \({4 \choose 1} \times\big(\frac{3}{4})^{10}\), likewise the probability that exactly two and three teams would be without customers are \({4 \choose 2} \time...

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_...

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...

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...

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 p...