Question

A martial artist is practicing breaking 5 boards. He is able to break aboard with probability 0.8.

(a) What is the probability that he breaks exactly 2 out of the 5 boards that are placed before him?

(b) What is the probability that he breaks at most 3 out of the 5 boards that are placed before him?

(c) What is the expected number of boards he will break?

Answer 1

a)
Here, n = 5, p = 0.8, (1 - p) = 0.2 and x = 2
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X = 2)
P(X = 2) = 5C2 * 0.8^2 * 0.2^3
P(X = 2) = 0.0512

b)
We need to calculate P(X <= 3).
P(X <= 3) = (5C0 * 0.8^0 * 0.2^5) + (5C1 * 0.8^1 * 0.2^4) + (5C2 * 0.8^2 * 0.2^3) + (5C3 * 0.8^3 * 0.2^2)
P(X <= 3) = 0.0003 + 0.0064 + 0.0512 + 0.2048
P(X <= 3) = 0.2627

c)
expected number = 5 * 0.8 = 4