probability.....yet again. (1 Viewer)

[kpq_sniper017]

Twelve students are to be chosen from twenty students of equal ability. Find the probability that:
(i) Three particular students A, B and C are chosen.
(ii) Students A and B are chosen but student C is not chosen.
(iii) None out of students A, B or C are chosen. A:14/285
(iv) At least one of the students A, B or C are chosen. A:271/285

I got (i) and (ii). I didn't get (iii) so I couldn't get (iv).

BTW. Say for instance (iv) was worth 2 marks. Would I get 1 mark just for saying that
P(at least one student being chosen) = 1 - P(none of them being chosen)
or would I have to do a bit more working?

 

[withoutaface]

i)17C9/20C12 since you have 3 automatically chosen, hence 9 places left and 17 people to fill them
ii)17C10/20C12 you have 2 auto. chosen, and one is removed altogether, hence 10 spots and 17 people to fill them
iii)17C12/20C12 You remove 3 from the mix, hence all the spots remain, but only 17 possible people can fill them
iv)1-answer iii), since the only criteria where this does not hold true is where none of them are chosen.

Hopefully this made sense, and is all correct. If you have a calculator near you perhaps check my results.

 

[kpq_sniper017]

ur answers r correct.
i didn't think of using combinations, coz this was on a 2U paper.

for (i) i said:
P=12/20 x 11/19 x 10/18 and it came up with the same result.

however, for (iii) i said:
P=8/20 x 8/19 x 8/18 (which was wrong) coz i thought it was the probability of not being chosen for each i.e. 1-12/20=8/20 for the 1st one.

what u said makes sense and they are right, so i think i'll stick to ur solutions
thx.