Question

Of the 20 flash bulbs in a box, 2 are defective. an inspector will select 2 light bulbs simultaneously and at random from the box. what is the probability that neither of the light bulbs selected will be defective

Answer 1

$|\Omega|=C(20,2)=\dfrac{20!}{2!18!}=\dfrac{19\cdot20}{2}=190\\ |A|=C(18,2)=\dfrac{18!}{2!16!}=\dfrac{17\cdot18}{2}=153\\\\ P(A)=\dfrac{153}{190}\approx81\%$