How Do You Solve a Venn Diagram Problem Involving Set Operations?

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Discussion Overview

The discussion revolves around solving a Venn diagram problem involving set operations, specifically focusing on probabilities related to students studying economics and history. Participants explore various calculations and interpretations of the problem, including the intersection and union of sets.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates the intersection of sets E and H, arriving at values for a, b, and c based on given equations.
  • Another participant confirms similar results using a system of equations derived from the problem statement.
  • Participants present probabilities related to students studying economics, with one stating the probability of a student not studying economics and suggesting cubing this result.
  • There is a disagreement regarding the interpretation of probabilities, particularly in part (ii) of the problem, with one participant asserting that the calculation leads to a probability exceeding 100%.
  • Another participant suggests that parts (i) and (ii) are mutually independent and that calculations for part (ii) can be derived from part (i).

Areas of Agreement / Disagreement

Participants generally agree on the calculations for part (i) but disagree on the interpretation and calculations for part (ii), with multiple competing views on how to approach the problem.

Contextual Notes

Some assumptions regarding the independence of events and the definitions of probabilities are not fully resolved, leading to different interpretations of the results.

karush
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(a) $E\cup H = 88-39=49$ and since $32+28=60$ then $b=60-49= E\cap H = 11$
so $a=32-11=21$ and $c=28-11=17$

hope this ok before (b) and (c)
 
Last edited:
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I get the same results as you. I wrote:

$$a+b=32$$

$$b+c=28$$

$$a+b+c=49$$

and solved the system.
 
(b)(I) $\frac{11}{88}=\frac{1}{8}$
(b)(ii) $\frac{56}{88}=\frac{7}{11}$
 
i) Correct.

ii) Incorrect. Given that he studies economics means the denominator is $a+b=32$. If he does not study history, then the numerator is $a=21$.
 
c)

i) Find the probability that one student does not study economics and then cube the result
ii) Note that this is mutually independent from part i).
 
eddybob123 said:
c)
1(i) Find the probability that one student does not study economics and then cube the result ii) Note that this is mutually independent from part i).
(i) $\displaystyle(\frac{56}{88})^3=\frac{343}{1331}$
 
karush said:
(i) $\displaystyle(\frac{56}{88})^3=\frac{343}{1331}$
(Clapping)

Did you do part ii)?
 
eddybob123 said:
(Clapping)

Did you do part ii)?

(C)(ii) in that the probability of just one student to take economics is $\displaystyle\frac{32}{88}$ i would presume that since $3$ students are randomly picked that $3x$ this would be the probability for at least one of these students to be in the econ class which would be $\displaystyle\frac{96}{88}$ which is more than a $100\%$

I was trying to do this with a cell phone yesterday and it took forever...but now on PC
 
Last edited:
karush said:
(C)(ii) in that the probability of just one student to take economics is $\displaystyle\frac{32}{88}$ i would presume that since $3$ students are randomly picked that $3x$ this would be the probability for at least one of these students to be in the econ class which would be $\displaystyle\frac{96}{88}$ which is more than a $100\%$

I was trying to do this with a cell phone yesterday and it took forever...but now on PC
i) and ii) are mutually independent and together represents the whole sample space. You don't need to do calculations to part ii) separately. Just take the result from i) and subtract it from 1.
 
  • #10
eddybob123 said:
i) and ii) are mutually independent and together represents the whole sample space. You don't need to do calculations to part ii) separately. Just take the result from i) and subtract it from 1.

$\frac{988}{1331}$
 

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