Container 1 has 8 items, 3 of which are defective. Container 2 has 5 items, 2 of which are defective. If one item is drawn from each container, what is the probability that only one of the items is defective?

Answer:The required probability is [tex]\frac{19}{40}[/tex]Step-by-step explanation:The probability of obtaining a defective item from container 1 is [tex]P(E_1)=\frac{3}{8}[/tex]The probability of obtaining a good item from container 1 is [tex]P(E_1)=\frac{5}{8}[/tex]The probability of obtaining a defective item from container 2 is [tex]P(E_1)=\frac{2}{5}[/tex]The probability of obtaining a good item from container 2 is [tex]P(E_1)=\frac{3}{5}[/tex]The cases of the event are1)Defective item is drawn from container 1 and good item is drawn from container 22)Defective item is drawn from container 2 and good item is drawn from container 1Thus the required probability is the sum of above 2 cases [tex]P(Event)=\frac{3}{8}\times \frac{3}{5}+\frac{5}{8}\times \frac{2}{5}=\frac{19}{40}[/tex]