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 advanced mathematics to solve.





This is a book on algorithms, some of them are probabilistic. But the book is a must have for students, job candidates even full time engineers & data scientists

















Good read. Overall Poker/Blackjack type card games are a good way to get introduced to probability theory





An excellent resource (students/engineers/entrepreneurs) if you are looking for some code that you can take and implement directly on the job.





Understanding Probability: Chance Rules in Everyday Life A bit pricy when compared to the first one, but I like the look and feel of the text used. It is simple to read and understand which is vital especially if you are trying to get into the subject





Data Mining: Practical Machine Learning Tools and Techniques, Third Edition (The Morgan Kaufmann Series in Data Management Systems) This one is a must have if you want to learn machine learning. The book is beautifully written and ideal for the engineer/student who doesn't want to get too much into the details of a machine learned approach but wants a working knowledge of it. There are some great examples and test data in the text book too.





This is a good book if you are new to statistics & probability while simultaneously getting started with a programming language. The book supports R and is written in a casual humorous way making it an easy read. Great for beginners. Some of the data on the companion website could be missing.

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?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