What is #P(x)# for #n=3. x=1, p=0.9#?

Question

3000 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-11T09:53:56

# p(1)=0.027#.

For a Binomial Random Variable #X# with parameters

#n and p,# then,

#P(X=x)=p(x)=""_nC_xp^nq^(n-x), x=0,1,2,...,n.#

Here, #n=3, p=0.9, x=1.#

Also, #q=1-p=1-0.9=0.1.#

#:. p(1)=""_3C_1(0.9)^1(0.1)^(3-1),#

#=3(0.9)(0.1)^2,#

#=3(0.9)(0.01).#

# rArr p(1)=0.027#.