Question

A carton contains 12 light bulbs of which 3 are defective. if a random sample of 4 light bulbs is chosen without replacement, what is the probability distribution of defective light bulbs?

Answer 1

\[P(0\ defective)=\frac{3C0\times9C4}{12C4}\]
\[P(1\ defective)=\frac{3C1\times9C3}{12C4}\]
\[P(2\ defective)=\frac{3C2\times9C2}{12C4}\]
\[P(3\ defective)=\frac{3C3\times9C1}{12C4}\]

Answer 2

Ahhhhh!

Answer 3

Thank u <3

Answer 4

You're welcome :)