PLEASE EXPLAIN REASONING/ASSUMPTIONS/STEPS. Thank you! A number x is selected at

Question

PLEASE EXPLAIN REASONING/ASSUMPTIONS/STEPS. Thank you!

A number x is selected at random in the interval [-1,2]. Let the events A = {x<0}, B = { |x-0.5| } < 0.5, and C = {x>0.75}.
(B is absolute value)
a. Find the probabilities of A, B, AnB, and AuC. (n represents intersection, u represents union)

b. Find the probabilities of AuB, AuC, and AuBuC,
first, by directly evaluating the sets and then their probabilities, and
second by using the appropriate axioms or corollaries.

Explanation / Answer

from above A will have values -1<X<0

B will have values -0.5<x-0.5<0.5 or 0<x<1

C will have values 0.75<x<2

as it is a uniform distribution function hence probabilty mass function f(X) =1/(2-(-1)) =1/3

a) probabilty of A: P(A) =P(-1<X<0) =(0-(-1))/3 =1/3

probabilty of B : P(B) =P(0<X<1) =(1-0)/3 =1/3

as A and B contain range -1 to 1

hence P(AnB) =P(-1<X<!) =(1-(-1))/3 =2/3

P(AUC) =P(-1<X<0)+P(0.75<X<2) =(0-(-1))/3+(2-0.75)/3 =1/3+1.25/3 =2.25/3 =9/12=3/4

b)P(AnB) =0; as they do not contain any common values in their domain.

P(AuB) =P(A)+P(B)-P(AnB) =(1/3)+(1/3)-0 =2/3

P(AnC) =0; as they do not contain any common values in their domain.

P(AUC)=P(A)+P(C)-P(AnC) =(1/3)+(1.25)/3-0 =2.25/3 =3/4

as B and C share 0.75<X<1 domain in their range

P(BnC) =0.25/3

hence P(AuBuC) =P(A)+P(B)+P(C) -P(AnB)-P(AnC)-P(BnC) +P(AnBnC)

=(1/3)+(1/3)+1.25/3-0-0-0.25/3+0

=1

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