Mathematics: Consider three events a, b and c with following properties. p (a) =1/4, p (b) = 1/6, p (c) = 1/4 p (aոb) = p (bոc) = p (aոc) = 1/9 p (aոbոc) = 1/13 let d = p (aubuc) c, find p (d). do the events a, b and c constitute the sample space? briefly justify your answer.

Consider three events a, b and c with following properties.

Ответ:

2020davidhines

11.01.2022

By the inclusion/exclusion principle,

$\mathbb P(D)=\mathbb P(A\cup B\cup C)$
$\mathbb P(D)=\mathbb P(A)+\mathbb P(B)+\mathbb P(C)-\bigg(\mathbb P(A\cap B)+\mathbb P(A\cap C)+\mathbb P(B\cap C)\bigg)+\mathbb P(A\cap B\cap C)$
$\mathbb P(D)=\dfrac14+\dfrac16+\dfrac14-\dfrac39+\dfrac1{13}$
$\mathbb P(D)=\dfrac{16}{39}\neq1$

So the union of A, B, and C does not constitute the entire sample space.

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Ответ:

MaeWolfe

23.05.2022

Mathematics

$1500!

20% off of $1500 would be $300. $1500 - $300 = $1200. Therefore the original sales price was $1500!

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Consider three events a, b and c with following properties.
p (a) =1/4, p (b) = 1/6, p (c) = 1/4
p (aոb) = p (bոc) = p (aոc) = 1/9
p (aոbոc) = 1/13
let d = p (aubuc) c, find p (d). do the events a, b and c constitute the sample space? briefly justify your answer.

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