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

[karush]

IB24
1202
2000
(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)

 

[MarkFL]

I get the same results as you. I wrote:

\(\displaystyle a+b=32\)

\(\displaystyle b+c=28\)

\(\displaystyle a+b+c=49\)

and solved the system.

 

[karush]

(b)(I) $\frac{11}{88}=\frac{1}{8}$
(b)(ii) $\frac{56}{88}=\frac{7}{11}$

 

[MarkFL]

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 $a=21$.

 

[eddybob123]

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).

 

[karush]

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) $\displaystyle(\frac{56}{88})^3=\frac{343}{1331}$

 

[eddybob123]

(i) $\displaystyle(\frac{56}{88})^3=\frac{343}{1331}$

(Clapping)

Did you do part ii)?

 

[karush]

(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

 

[eddybob123]

(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.

 

[karush]

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.

$\frac{988}{1331}$