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jeka94
3 years ago
8

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?
Mathematics
1 answer:
e-lub [12.9K]3 years ago
4 0

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}

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