Respuesta :

LammettHash
By the inclusion/exclusion principle,

[tex]\mathbb P(E\cup F)=\mathbb P(E)+\mathbb P(F)-\mathbb P(E\cap F)[/tex]

Recall the law of total probability:

[tex]\mathbb P(F)=\mathbb P(E\cap F)+\mathbb P(E^C\cap F)[/tex]

We're given

[tex]P(F^C)=\dfrac7{20}\implies\mathbb P(F)=1-\mathbb P(F^C)=\dfrac{13}{20}[/tex]

and so

[tex]\dfrac{13}{20}=\mathbb P(E\cap F)+\dfrac{31}{100}\implies\mathbb P(E\cap F)=\dfrac{17}{50}[/tex]

Now,

[tex]\mathbb P(E\cup F)=\dfrac{57}{100}+\dfrac{13}{20}-\dfrac{17}{50}=\dfrac{22}{25}[/tex]
MrRoyal

Probabilities are used to determine the chances of an event.

The value of P(e u f) is 22/25.

The given parameters are:

[tex]\mathbf{P(e) = \frac{57}{100}}[/tex]

[tex]\mathbf{P(f^c) = \frac{7}{20}}[/tex]

[tex]\mathbf{P(f\ n\ e^c) = \frac{31}{100}}[/tex]

First, we calculate P(f).

Using the complement rule, we have:

[tex]\mathbf{P(f) = 1 - P(f^c)}[/tex]

[tex]\mathbf{P(f) =1 - \frac{7}{20}}[/tex]

[tex]\mathbf{P(f) =\frac{13}{20}}[/tex]

Also, we have:

[tex]\mathbf{P(f\ n\ e^c) = \frac{31}{100}}[/tex]

This gives:

[tex]\mathbf{P(f\ n\ e^c) = P(f) - P(f\ n\ e)}[/tex]

So, we have:

[tex]\mathbf{P(f\ n\ e) = P(f) - P(f\ n\ e^c)}[/tex]

[tex]\mathbf{P(f\ n\ e) = \frac{13}{20} - \frac{31}{100}}[/tex]

[tex]\mathbf{P(f\ n\ e) = \frac{65 - 31}{100}}[/tex]

[tex]\mathbf{P(f\ n\ e) = \frac{34}{100}}[/tex]

So, we have:

[tex]\mathbf{P(e\ u\ f) = P(e) + P(f) - P(f\ n\ e)}[/tex]

This gives

[tex]\mathbf{P(e\ u\ f) = \frac{57}{100} + \frac{13}{20} - \frac{34}{100}}[/tex]

Take LCM

[tex]\mathbf{P(e\ u\ f) = \frac{57 + 65 - 34}{100}}[/tex]

[tex]\mathbf{P(e\ u\ f) = \frac{88}{100}}[/tex]

Simplify

[tex]\mathbf{P(e\ u\ f) = \frac{22}{25}}[/tex]

Hence, the value of P(e u f) is 22/25.

Read more about probabilities at:

https://brainly.com/question/11234923

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