Answer:
<h3>For two events A and B show that P (A∩B) ≥ P (A)+P (B)−1.</h3>
By De morgan's law

which is Bonferroni’s inequality
<h3>Result 1: P (Ac) = 1 − P(A)</h3>
Proof
If S is universal set then

<h3>Result 2 : For any two events A and B, P (A∪B) = P (A)+P (B)−P (A∩B) and P(A) ≥ P(B)</h3>
Proof:
If S is a universal set then:

Which show A∪B can be expressed as union of two disjoint sets.
If A and (B∩Ac) are two disjoint sets then
B can be expressed as:

If B is intersection of two disjoint sets then

Then (1) becomes

<h3>Result 3: For any two events A and B, P(A) = P(A ∩ B) + P (A ∩ Bc)</h3>
Proof:
If A and B are two disjoint sets then

<h3>Result 4: If B ⊂ A, then A∩B = B. Therefore P (A)−P (B) = P (A ∩ Bc) </h3>
Proof:
If B is subset of A then all elements of B lie in A so A ∩ B =B
where A and A ∩ Bc are disjoint.

From axiom P(E)≥0

Therefore,
P(A)≥P(B)
Answer:
The change in the standard of living, measured by growth in real GDP per person, is 0.1356%
Step-by-step explanation:
Given,
In the current year,
Population = 31.5 million,
Real GDP = $814 million,
So, the real GDP per person =
≈ 25.841
In the previous year,
Population = 31 million,
Real GDP = $800 million,
So, the real GDP per person =
≈ 25.806
Hence, change percentage = 



≈ 0.1356 %
Answer:
I dont think it can be simplified
Step-by-step explanation:
Answer:
If AB = CD, then CD = AB. symmetric property
If MN = XY , and XY = AB , then MN = AB. transitive property
Segment CD is congruent to segment CD reflexive property
Step-by-step explanation:
I just finished the assignment
Answer:
which question? you want me to answer
Step-by-step explanation: