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

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SUMMARY

The discussion focuses on the analysis of Venn diagrams, specifically the union and intersection of events A and B within a sample space U. It establishes that if the complement of the union of A and B is 21 and the total number of elements in U is 36, then the number of elements in the union of A and B is 15, leading to an intersection of 2 elements. The probability of the intersection, P(A∩B), is calculated as 1/18. Additionally, the concept of mutually exclusive sets is clarified, indicating that two sets are mutually exclusive if their intersection is empty.

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  • Familiarity with set theory concepts such as union and intersection
  • Basic knowledge of probability calculations
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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..
 
Last edited:
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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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