Question

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 1

Answer:

The required probability is $\frac{19}{40}$

Step-by-step explanation:

The probability of obtaining a defective item from container 1 is $P(E_1)=\frac{3}{8}$

The probability of obtaining a good item from container 1 is $P(E_1)=\frac{5}{8}$

The probability of obtaining a defective item from container 2 is $P(E_1)=\frac{2}{5}$

The probability of obtaining a good item from container 2 is $P(E_1)=\frac{3}{5}$

The cases of the event are

1)Defective item is drawn from container 1 and good item is drawn from container 2

2)Defective item is drawn from container 2 and good item is drawn from container 1

Thus the required probability is the sum of above 2 cases

$P(Event)=\frac{3}{8}\times \frac{3}{5}+\frac{5}{8}\times \frac{2}{5}=\frac{19}{40}$