Answer to Question #332680 in Statistics and Probability for jm2

Question

two incandescent lights are chosen at random from 19 lights of which 4 are defective. Find the probability that (a) none is defective? (b) exactly one is defective?

Answer 1

There are $C_{19}^2=\frac{18\cdot19}{2}=9\cdot19=171$ ways to choose $2$ lights from $19$. a). There are $C_{15}^2=\frac{14\cdot15}{2}=7\cdot 15=105$ ways to choose $2$ lights from $15$ lights without defects. Thus, the probability is given by $p_1=\frac{C_{15}^2}{C_{19}^2}=\frac{105}{171}\approx0.614.$ b). There are $C_{15}^1=15$ ways to choose $1$ light without defect and $C_4^1=4$ ways to choose $1$ light with defect. Thus, the probability is as follows: $p_2=\frac{C_{15}^1C_4^1}{C_{19}^2}=\frac{60}{171}\approx0.351$.

The answers are as follows: a) $p_1\approx0.614$, b) $p_2\approx0.351.$