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[SOLVED] IB24 represented in the following Venn diagram

karush

Well-known member
Jan 31, 2012
2,725
View attachment 1202
(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:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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

Well-known member
Jan 31, 2012
2,725
(b)(I) $\frac{11}{88}=\frac{1}{8}$
(b)(ii) $\frac{56}{88}=\frac{7}{11}$
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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

Active member
Aug 18, 2013
76
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

Well-known member
Jan 31, 2012
2,725
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

Active member
Aug 18, 2013
76

karush

Well-known member
Jan 31, 2012
2,725
(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:

eddybob123

Active member
Aug 18, 2013
76
(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

Well-known member
Jan 31, 2012
2,725
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.
$\frac{988}{1331}$