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OpenStudy (helpmeplease14):
OpenStudy (xapproachesinfinity):
so let's see here
we have to openings to be filled yes?
OpenStudy (xapproachesinfinity):
two*
OpenStudy (helpmeplease14):
?
OpenStudy (xapproachesinfinity):
I'am asking yo do you agree that there are only two spots to fill by two candidates
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OpenStudy (helpmeplease14):
o oh yes
OpenStudy (xapproachesinfinity):
good! now do you agree that we need to choose two out of 4?
OpenStudy (helpmeplease14):
But theres 5 students
OpenStudy (xapproachesinfinity):
oh sorry 2 out of 5 :) see I'm making you think lol
OpenStudy (helpmeplease14):
Lol
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OpenStudy (xapproachesinfinity):
now I'am sure you are following hehe
OpenStudy (helpmeplease14):
Yeah
OpenStudy (xapproachesinfinity):
so we have 5 choose 2 correct?
OpenStudy (xapproachesinfinity):
that would give us the chances that two are selected!
OpenStudy (xapproachesinfinity):
assuming that all candidates have equal chance to be selected!
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OpenStudy (xapproachesinfinity):
so do you know how to compute 5 choose 2?
OpenStudy (helpmeplease14):
Compute?
OpenStudy (xapproachesinfinity):
yes! you should have this formula \[\Large C_{2}^{5} \]
OpenStudy (helpmeplease14):
?
OpenStudy (helpmeplease14):
I didn't learn that yet
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OpenStudy (xapproachesinfinity):
\[C_{k}^{n}=\frac{n!}{k!(n-k)!}\]
you sure you didn't learn this
OpenStudy (helpmeplease14):
Yeah
OpenStudy (xapproachesinfinity):
you did it or not?
OpenStudy (helpmeplease14):
I never done that
OpenStudy (helpmeplease14):
Is there another way?
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OpenStudy (xapproachesinfinity):
hmm this changes the whole situation then
OpenStudy (xapproachesinfinity):
hmm don't know what did you learn so far,
did you learn any arrangement method?
OpenStudy (helpmeplease14):
No I'm in understanding probability of compound events
OpenStudy (xapproachesinfinity):
compound events?
OpenStudy (helpmeplease14):
Yeah
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OpenStudy (xapproachesinfinity):
i'm asking what do you mean by compound events?
OpenStudy (xapproachesinfinity):
never hear that in probs
OpenStudy (helpmeplease14):
That's the name of the lesson I'm doing
OpenStudy (xapproachesinfinity):
hmm anyways let me see how else we can do this
OpenStudy (helpmeplease14):
K
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OpenStudy (xapproachesinfinity):
P(Emma and a boy)=P(Emma)p(a boy)
p(emma)=1/5 and P(a boy)=3/5
is besty a name of a girl?
OpenStudy (xapproachesinfinity):
did you do \[P(A\cap B)\]
when A and B are independent events or not
OpenStudy (helpmeplease14):
Yeah I'm learning that! :)
OpenStudy (xapproachesinfinity):
okay so we are dealing with dependent events yes?
OpenStudy (helpmeplease14):
Yes and I got 12
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OpenStudy (xapproachesinfinity):
12 what?
OpenStudy (xapproachesinfinity):
probability is never bigger than 1 you know
OpenStudy (xapproachesinfinity):
what is the p(Emma)?
OpenStudy (helpmeplease14):
1/5
OpenStudy (xapproachesinfinity):
and P(a boy is chosen)=?
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OpenStudy (xapproachesinfinity):
i assumed above that besty is a girl not sure about the name lol
OpenStudy (helpmeplease14):
3/5
OpenStudy (xapproachesinfinity):
then 1/5 times 3/5 is what?
OpenStudy (helpmeplease14):
3/25
OpenStudy (xapproachesinfinity):
good! that should be the first
now prob that emma is not choosen
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OpenStudy (xapproachesinfinity):
let me see you do that one
OpenStudy (helpmeplease14):
3/4 × 1/4 = 4/16 = 1/4
OpenStudy (xapproachesinfinity):
hmm can't see what you did
use the drawing box
OpenStudy (helpmeplease14):
I can't
OpenStudy (xapproachesinfinity):
well anyway this what you need to use
\[P(A^c)=1-P(A)\]
where \[A^c\] is the complement of A
for our problem A is choosing Emma
P(Emma)=1/5
so P( not choosing Emma)=1-P(Emma)
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