Question

The probability that a patients recovers from a rare blood disease is ‘0.55’. If 15

people are known to have contracted this disease. What is the probability that from 3

to ‘55’ survive?

Answer 1

Let $X=$ the number of patients who survived: $X\sim Bin(n, p).$

Given $n=15, p=0.55, q=1-p=0.45.$

$$P(X=3)=\dbinom{15}{3}(0.55)^3(0.45)^{15-3}\approx0.00521975$$