Please help me asap!

For two events A and B show that P (A∩B) ≥ P (A)+P (B)−1.

nordsb [41]

Answer:

<h3>For two events A and B show that P (A∩B) ≥ P (A)+P (B)−1.</h3>

By De morgan's law

$(A\cap B)^{c}=A^{c}\cup B^{c}\\\\P((A\cap B)^{c})=P(A^{c}\cup B^{c})\leq P(A^{c})+P(B^{c}) \\\\1-P(A\cap B)\leq P(A^{c})+P(B^{c}) \\\\1-P(A\cap B)\leq 1-P(A)+1-P(B)\\\\-P(A\cap B)\leq 1-P(A)-P(B)\\\\P(A\cap B)\geq P(A)+P(B)-1$

which is Bonferroni’s inequality

<h3>Result 1: P (Ac) = 1 − P(A)</h3>

Proof

If S is universal set then

$A\cup A^{c}=S\\\\P(A\cup A^{c})=P(S)\\\\P(A)+P(A^{c})=1\\\\P(A^{c})=1-P(A)$

<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:

$A\cup(B\cap A^{c})=(A\cup B) \cap (A\cup A^{c})\\=(A\cup B) \cap S\\A\cup(B\cap A^{c})=(A\cup B)$

Which show A∪B can be expressed as union of two disjoint sets.

If A and (B∩Ac) are two disjoint sets then

$P(A\cup B) =P(A) + P(B\cap A^{c})---(1)\\$

B can be expressed as:

$B=B\cap(A\cup A^{c})\\$

If B is intersection of two disjoint sets then

$P(B)=P(B\cap(A)+P(B\cup A^{c})\\P(B\cup A^{c}=P(B)-P(B\cap A)$

Then (1) becomes

$P(A\cup B) =P(A) +P(B)-P(A\cap B)\\$

<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

$A=A\cap(B\cup B^{c})\\A=(A\cap B) \cup (A\cap B^{c})\\P(A)=P(A\cap B) + P(A\cap B^{c})\\$

<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

$A =(A \cap B)\cup (A\cap B^{c}) = B \cup ( A\cap B^{c})$

where A and A ∩ Bc are disjoint.

$P(A)=P(B\cup ( A\cap B^{c}))\\\\P(A)=P(B)+P( A\cap B^{c})$

From axiom P(E)≥0

$P( A\cap B^{c})\geq 0\\\\P(A)-P(B)=P( A\cap B^{c})\\P(A)=P(B)+P(A\cap B^{c})\geq P(B)$

Therefore,

P(A)≥P(B)

8 0

2 years ago

A Number is multiplied by itself. The product is 5 4/9 Find the number

Over [174]

Answer:

it is 2/3

Step-by-step explanation

i can't explain it to you

8 0

2 years ago

Read 2 more answers

can someone help me figure this out?? My sisters at work rn she usually helps me with algerbra pleasee help!

ANEK [815]

Answer:

its the last one hun

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

if your bicycle tire has a diameter of 9 inches. how many rotations does each tire make if you travel 300 feet?

Vesnalui [34]

Step One
Calculate the number of feet traveled in 1 rotation.

Formula
C = π*d
P = 3.14
d = 9 inches = 9/12 feet = 3/4 of a foot.

C = 3.14 * 3/4 = 2.355 So that means that every time the tire turns around 1 complete turn, the distance traveled on the ground is 2.355 feet.

Step Two
Figure out the number of revolutions.
1 revolution = 2.355 feet
x revolutions = 300 feet.

1/x = 2.355/ 300 Cross multiply
2.355 feet * x = 1 rev * 300 feet
2.355 x = 300 rev Divide by 2.355
x = 300 / 2.355
x = 127.39 revolutions. <<<< Answer

8 0

3 years ago

Deer ticks can be carriers of either Lyme disease or human granulocytic ehrlichiosis (HGE). Based on a recent study, suppose tha

san4es73 [151]

Answer:

0.2364

Step-by-step explanation:

We will take

Lyme = L

HGE = H

P(L) = 16% = 0.16

P(H) = 10% = 0.10

P(L ∩ H) = 0.10 x p(L U H)

Using the addition theorem

P(L U H) = p(L) + P(H) - P(L ∩ H)

P(L U H) = 0.16 + 0.10 - 0.10 * p(L u H)

P(L U H) = 0.26 - 0.10p(L u H)

We collect like terms

P(L U H) + 0.10P(L U H) = 0.26

This can be rewritten as:

P(L U H)[1 +0.1] = 0.26

Then we have,

1.1p(L U H) = 0.26

We divide through by 1.1

P(L U H) = 0.26/1.1

= 0.2364

Therefore

P(L ∩ H) = 0.10 x 0.2364

The probability of tick also carrying lyme disease

P(L|H) = p(L ∩ H)/P(H)

= 0.1x0.2364/0.1

= 0.2364

3 0

2 years ago