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

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The discussion focuses on solving a Venn diagram problem involving set operations with specific probabilities related to students studying economics and history. The calculations reveal that the intersection of students studying both subjects is 11, while the individual counts are 21 for economics and 17 for history. The probability of a student not studying economics is calculated as \((\frac{56}{88})^3 = \frac{343}{1331}\), and the probability of at least one student studying economics among three randomly chosen students is incorrectly presumed to be \(\frac{96}{88}\), exceeding 100%. The participants emphasize the independence of the events and the importance of using the correct sample space.

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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.
 
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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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