Answer:
Step-by-step explanation:
Given that there are 20 non ionized containers and 73 ionized containers
Two samples are drawn without replacement
a) the probability that the first one selected is not ionized=[tex]\frac{20}{93} =0.215[/tex]
b) the probability that the second one selected is not ionized given that the first one was ionized
= When first one was ionized we got left over as 20 and 72
Hence = [tex]\frac{20}{92} =0.217[/tex]
c) If with replacement left over 20 and 73 and hence prob = 0.215 as in part a
d) the probability that both are ionized=[tex]\frac{73C2}{93C2} =0.614[/tex]
