If two orders are selected, find the probability that they are both accurate. Complete parts (a) and (b) below

A. Assume that the selections are made with replacement. Are the events independent?

Respuesta :

Probabilities are used to determine the chances of events

  • The probability with replacement is 0.7742, while the probability without replacement is 0.7741
  • The probability with replacement is independent, while the probability without replacement is dependent

From the table (see attachment), we have:

[tex]\mathbf{Order\ Accurate = 331 + 269 + 245 +144}[/tex]

[tex]\mathbf{Order\ Accurate = 989}[/tex]

[tex]\mathbf{Order\ Not\ Accurate = 34 + 53 + 30 + 18}[/tex]

[tex]\mathbf{Order\ Not\ Accurate = 135}[/tex]

(a) Probability that both orders are accurate (with replacement)

This is calculated using:

[tex]\mathbf{Pr = \frac{Order\ Accurate}{Total} \times \frac{Order\ Accurate}{Total}}[/tex]

So, we have:

[tex]\mathbf{Pr = \frac{989}{989+135} \times \frac{989}{989 + 135}}[/tex]

[tex]\mathbf{Pr = \frac{989}{1124} \times \frac{989}{1124}}[/tex]

[tex]\mathbf{Pr = 0.7742}[/tex]

The events are independent, because the first selection does not affect the second

(b) Probability that both orders are accurate (without replacement)

This is calculated using:

[tex]\mathbf{Pr = \frac{Order\ Accurate}{Total} \times \frac{Order\ Accurate - 1}{Total - 1}}[/tex]

So, we have:

[tex]\mathbf{Pr = \frac{989}{989+135} \times \frac{989 - 1}{989 + 135 - 1}}[/tex]

[tex]\mathbf{Pr = \frac{989}{1124} \times \frac{988}{1123}}[/tex]

[tex]\mathbf{Pr = 0.7741}[/tex]

The events are dependent, because the first selection affects the second

Read more about probabilities at:

https://brainly.com/question/11234923

Ver imagen MrRoyal