Question

The black casing for a model of laptop is to be made from a large bag of mixed plastic pellets (ie very small balls), 97% of the pellets are the raw plastic, 3% of the pellets are the black pigment. A person randomly picked 4 pellets without replacement, find the probability of getting 3 pellets of raw plastic and 1 pellet of the black pigment.

Answer 1

Let $X=$ the number of raw plastic pigment:$X\sim Bin(n, p)$

$$P(X=x)=\dbinom{n}{x}p^x(1-p)^{n-x}$$

Given $p=0.97, n=4$

$$P(X=3)=\dbinom{4}{3}(0.97)^3(1-0.97)^{4-3}$$

$$=4(0.97)^3(0.03)=0.10952076$$