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

The required probability is
`(19)/(40)`

Explanation:

The probability of obtaining a defective item from container 1 is
`P(E_1)=(3)/(8)`

The probability of obtaining a good item from container 1 is
`P(E_1)=(5)/(8)`

The probability of obtaining a defective item from container 2 is
`P(E_1)=(2)/(5)`

The probability of obtaining a good item from container 2 is
`P(E_1)=(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)=(3)/(8)* (3)/(5)+(5)/(8)* (2)/(5)=(19)/(40)`