MHB -aux.02 Venn diagram sample space U and events A and B

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The discussion focuses on analyzing a Venn diagram involving sample space U and events A and B. It confirms that the shaded area represents the complement of the union of A and B, with calculations showing that if the complement has 21 elements and the total sample space has 36, then the union of A and B contains 15 elements. The intersection of A and B is determined to be 2, leading to a probability calculation for the intersection, which is clarified to be 1/18 after some corrections. Additionally, the concept of mutually exclusive sets is explained, indicating that they do not overlap.
karush
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(a) the shaded hopefully shows $(A\cup B)'$

(b) (i) if $(A\cup B)'= 21$ and $n(U)=36$ then $n(A\cup B)=15$
but $n(A)+n(B)=17$ so $n(A\cap B) = 2$

(ii) $P(A\cap B)$ not sure but guessing $2:17$

(c) not sure what "mutually exclusive" means but presume it has to do with the overlap.

never done Venn Diagrams so this is all new ... did look at De Morgan's stuff tho..
 
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(a) Correct. Anything that is not in either A or B should be shaded, which is what you did.

(b)

(i) Correct.

(ii) I believe you want to use:

$$P(A\cap B)=\frac{n(A\cap B)}{n(U)}$$

(c) Two sets are mutually exclusive if:

$$A\cap B=\emptyset$$
 
MarkFL said:
(a)

(ii) I believe you want to use:

$$P(A\cap B)=\frac{n(A\cap B)}{n(U)}$$

so

$\displaystyle P(A\cap B)=\frac{n(A\cap B)}{n(U)}=

\frac{2}{36}=\frac{1}{16}$
 
karush said:
so

$\displaystyle P(A\cap B)=\frac{n(A\cap B)}{n(U)}=

\frac{2}{36}=\frac{1}{16}$

Not quite...

$$\frac{2}{36}=\frac{2}{2\cdot18}=\frac{1}{18}$$

:D
 
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