Mathematics: 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?

Rewrite the fractions as top heavy:1 + 2/3 = 5 / 36 + 1/2 = 13 / 213 / 2 * 5 / 3 = 65 / 6 hours...

Container 1 has 8 items, 3 of which are defective.

heleneallie7ou1fnk

16.03.2022

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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