Binomial Board Breaking
Bruce Lee is practicing his skills for an unreleased movie by breaking 5 boards with his fists.
Assume he breaks a board with probability 0.8 (he is practicing his left fist; that's why the probability is not 1) and that he breaks each board independently.

What is the probability that Bruce breaks exactly 2 out of the 5 boards that are placed before him? Please answer in the form of a decimal with four significant digits.
This is the standard Binomial probability: There are \(\binom{5}{2}\) ways for 2 of the 5 boards to be broken. Each one of these ways happens with probability \(0.8^2\cdot 0.2^3\). Overall, the probability is: \[\binom{5}{2}\cdot 0.8^2\cdot 0.2^3 = \frac{5!}{3!\cdot 2!}\cdot\left(\frac{8}{10}\right)^2\cdot\left(\frac{2}{10}\right)^3\] 
What is the probability that Bruce breaks at most 3 out of the 5 boards that are placed before him? Please answer in the form of a decimal with five significant digits.
This is 1 minus the probability that 4 or 5 boards will be broken. There is 1 way for all 5 boards to be broken, and this happens with probability \(0.8^5\). Similarly, there are \(\binom{5}{4}\) ways for 4 boards to be broken, and each happens with probability \(0.8^4\cdot 0.2\). So, the overall probability is: \[1\binom{5}{4}\cdot 0.8^4\cdot 0.20.8^5\] 
What is the expected number of boards Bruce will break?
The expectation of a \((p,n)\) binomial distribution is \(np\). Here, \(p = 0.8\) and \(n = 5\).