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:
Correlation does not always imply causation. (It's not correct)
Step-by-step explanation:
Correlation tests 2 variables' relationship, but just because they are related does not mean they necessarily cause each other.
Answer:
24
Step-by-step explanation:
12 x 2
I believe it’s A, C, and E
Answer:
IBi0
Step-by-step explanation:
We look at the child's genotype (lAi0).
We know that the blood type's genotype has 2 genes, one from the mother and one from the father.
If the genotype is lAi0 and one gene is, let's say, from mother (lA), we know exactly that the other gene is from father (i0).
So, father's genotype is lBi0 (a).
