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Mathematics 25 Online
OpenStudy (helpmeplease14):

Can someone help me

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

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

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!

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

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?

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

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

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

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)=?

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

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)

OpenStudy (helpmeplease14):

1 × 1/5

OpenStudy (xapproachesinfinity):

yes

OpenStudy (helpmeplease14):

OK and I got 1/5

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