anonymous
  • anonymous
help me please.. Show that if P (A) >0 , then P(A ∩ B | A) ≥ P(A∩B| A ∪B )
Probability
katieb
  • katieb
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

Zarkon
  • Zarkon
use the definition of conditional probability and the fact that \[P(A)\le P(A\cup B)\]
anonymous
  • anonymous
okay i'll try first :)
anonymous
  • anonymous
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)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
is that true ?
Zarkon
  • Zarkon
yes

Looking for something else?

Not the answer you are looking for? Search for more explanations.