4.5 Expectation

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.

1. 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.

Exercise 1

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: $\binom{5}{2}\cdot 0.8^2\cdot 0.2^3 = \frac{5!}{3!\cdot 2!}\cdot\left(\frac{8}{10}\right)^2\cdot\left(\frac{2}{10}\right)^3$

2. 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.

Exercise 2

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: $1-\binom{5}{4}\cdot 0.8^4\cdot 0.2-0.8^5$

3. What is the expected number of boards Bruce will break?

Exercise 3

The expectation of a $$(p,n)$$- binomial distribution is $$np$$. Here, $$p = 0.8$$ and $$n = 5$$.