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A box contains 100 items of which 4 are defective. Two items are chosen at random from the box. What is the probability of selecting (a) 2 defectives if the first item is not replaced; (b) 2 defectives if the first item is put back before choosing the second item; (c) 1 defective and I non-defective if the first item is not replaced?

Respuesta :

A probability is an amount that reflects these same probabilities. Probabilities can be expressed in percentages from 0 to 1, as well as in proportions from 0 to 100 %.

Following are calculations of the given points:

A)

[tex]\to \bold{P(2 \ d efectives ) =\frac{4}{100}\times \frac{3}{99} =0.0012}}[/tex]  

B)

[tex]\to \bold{P(2\ d efectives)=\frac{4}{100}\times \frac{4}{100}=0.0016}[/tex]

C)

[tex]\to \bold{P(1 \ d efective)=\left (\frac{4}{100}\times \frac{96}{99} \right )+\left ( \frac{96}{100} \times\frac{4}{99} \right )=0.0776}[/tex]

So, the final answer is "0.0012, 0.0016, and 0.0776"

Learn more:

brainly.com/question/15056962

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