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LiRa [457]
3 years ago
15

If P(A) = 0.2, P(B) = 0.2, and P(A ∪ B) = 0.4, then P(A ∩ B) = _______.

Mathematics
1 answer:
Pavlova-9 [17]3 years ago
8 0

Answer:

P(A ∩ B) = 0.

a) NO

b) YES

Step-by-step explanation:

Thinking about this through Venn diagrams we can sort of understand that:

if P(A) = 0.2 and P(B) = 0.2, and P(A∪B) = 0.4.

there's no overlapping between P(A) and P(B).

(If there was overlapping then P(A∪B) < 0.4, since you'd be excluding the overlapped part from getting counted twice.

Think of it in terms of calculating areas circles A and B, if the circles were disjoint, then the sum of the areas A and B would be 0.2+0.2. But if the circles were overlapping then the sum of the areas would be 0.2+0.2-P(A ∩ B), where P(A ∩ B) is the overlapping part)

since there's no overlapping P(A ∩ B) = 0.

a) NO

events A and B are only independent when P(A ∩ B) > 0 (or overlapping)

b) YES

events A and B are mutually exclusive when P(A ∩ B) = 0 (or disjoint)

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The solution to the division of the given surd is: \mathbf{P =\dfrac{(6\sqrt{x}-x^2\sqrt{x}-3x\sqrt{x}+2x+2)(x-1) }{8x}      }

<h3>Division of Surds.</h3>

The division of surds follows a systemic approach whereby we divide the whole numbers separately and the root(s) are being divided by each other.

Given that:

\mathbf{P=(\frac{\sqrt{x} +2}{x-1}+\frac{\sqrt{x}\\ -2}{x-2\sqrt{x} +1} ) : \frac{4x}{(x-1)^{2} }}

i.e.

\mathbf{=\dfrac{(\frac{\sqrt{x} +2}{x-1}+\frac{\sqrt{x}\\ -2}{x-2\sqrt{x} +1} )}{ \frac{4x}{(x-1)^{2} }} }

Using the fraction rule:

\mathbf{\dfrac{a}{\dfrac{b}{c}}= \dfrac{a\times c}{b}}

\mathbf{\implies \dfrac{(\frac{\sqrt{x} +2}{x-1}+\frac{\sqrt{x}\\ -2}{x-2\sqrt{x} +1} )(x-1)^{2}}{4x}} }

By simplification, we have:

\mathbf{  =\dfrac{\dfrac{(6\sqrt{x}-x^2\sqrt{x}-3x\sqrt{x}+2x+2)(x-1) }{2}  }{4x}      }

\mathbf{P =\dfrac{(6\sqrt{x}-x^2\sqrt{x}-3x\sqrt{x}+2x+2)(x-1) }{8x}      }

Learn more about evaluating the division of surds here:

https://brainly.in/question/27942899

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Step-by-step explanation:


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Step-by-step explanation:

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