hypergeometric distribution In a quiz requiring Minitab in B

In a quiz requiring Minitab in BE 2100, 36 computers are provided, of which 8 do not have Minitab installed on them. The first 6 students that came in selected a computer each. What is the probability that at most one of them selected a computer without Minitab?

Solution

let X be the random variable denoting the number of students out of six students without having minitab in his computer.

there are all total 36 computers out of which 8 dont have minitab installed and there are 6 students.

so X~Hypergeometric(36,8,6)

so pmf of X is f(x)=⁸C_x*^36-8C_6-x/³⁶C₆     x=0,1,2,3,4,5,6

we are to find P[at most one them selected a computer without minitab]=P[X<=1]=P[X=0]+P[X=1]=⁸C₀*^36-8C_6-0/³⁶C₆+⁸C₁*^36-8C_6-1/³⁶C₆=0.597076 [answer]