Question

help me please.. Show that if P (A) >0 , then
P(A ∩ B | A) ≥ P(A∩B| A ∪B )

Answer 1

use the definition of conditional probability and the fact that

\[P(A)\le P(A\cup B)\]

Answer 2

okay i'll try first :)

Answer 3

P( A intersect B | A) = P( (A intersect B) intersect A)/P(A) = P(A intersect B)/P(A)
P( A intersect B | A union B) = P( {A intersect B} intersect {A union B})/P(A union B)
..= P(A intersect B)/P(A union B).

As the numerators for the two expressions are equal, while the denominator defining
P( A intersect B | A union B) is larger than the denominator defining P( A intersect B | A ),
it follows that P( A intersect B | A) >= P( A intersect B | A union B)

Answer 4

is that true ?

Answer 5

yes