Question

1- After a particularly hard surgery, probability that person will live is 0.9. What is the probability that

exactly 4 of the next 6 people having this operation survive.

2- A group of 3 people is selected from four professors and 2 students. What is the probability that there is

at most 1 student in this group?

Answer 1

1. Binomial distribution: P(X) = nCx p^x q^n-x

P(a person survives), p = 0.9

q = 1 - p = 0.10

n = 6

P(exactly 4 of the next 6 people having this operation survive) = 6C4 x 0.9⁴ x 0.1²

= 0.0984

2. Combination formula: nCr = n!/(r! x (n-r)!)

Number of professors = 4

Number of students = 2

P(at most 1 student) = (Number of ways to select 0 students and 3 professors + Number of ways to select 1 student and 2 professors) / Number of ways to select any 3 from 6

= (1 x 4C3 + 2C1 x 4C2)/6C3

= (1x4 + 2x6)/20

= 0.8