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23 Feb 2018, 08:37
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One option is to apply probability rules.
So, for two siblings to be selected, 2 things must happen: we must select a junior who has a senior sibling AND the senior selected must be the sibling of the selected junior.
We get P(junior with sibling AND selected senior is sibling to selected junior)= P(junior with sibling) x P(selected senior is sibling to selected junior)
P(junior with sibling): there are 1000 juniors and 60 of them have senior siblings. So, P(junior with sibling)=60/1000
P(selected senior is sibling to selected junior): Once the junior has been selected, there is only 1 senior (out of 800 seniors) who is the sibling to the selected junior. So, P(selected senior is sibling to selected junior)= 1/800
So, the probability is (60/1000)x(1/800) = 3/40,000
The answer is A
Cheers,
Brent
16 Sep 2018, 15:07
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\(\left( J \right)\,\,{\text{Junior}}\,\,:\,\,\,1000\,\,{\text{students}}\,\,,\,\,\,{J_1}\,,\,{J_2},\,\, \ldots \,\,,\,\,{J_{60}}\,\,{\text{among}}\,\,{\text{them}}\,\,\,\left( {{\text{the}}\,\,{\text{ones}}\,\,{\text{with}}\,\,{\text{senior}}\,\,{\text{siblings}}} \right)\)
\(\left( S \right)\,\,{\text{Senior}}\,\,:\,\,\,800\,\,{\text{students}}\,\,,\,\,\,{S_1}\,,\,{S_2},\,\, \ldots \,\,,\,\,{S_{60}}\,\,{\text{among}}\,\,{\text{them}}\,\,\,\left( {{\text{the}}\,\,{\text{ones}}\,\,{\text{with}}\,\,{\text{junior}}\,\,{\text{siblings}}} \right)\)
\(\left( {{J_1}\,,\,{S_1}} \right)\,\,;\,\,\left( {{J_2}\,,\,{S_2}} \right)\,\,;\,\, \ldots \,\,;\,\,\left( {{J_{60}}\,,\,{S_{60}}} \right)\,\,\,:\,\,\,{\text{pairs}}\,\,{\text{of}}\,\,{\text{siblings}}\,\)
\(? = P\left( {{\text{pair}}\,{\text{of}}\,{\text{siblings}}\,,\,\,{\text{in}}\,{\text{one}}\,J\,{\text{and}}\,{\text{one}}\,S\,{\text{extraction}}} \right)\)
\({\text{Total}}\,\,:\,\,\,1000 \cdot 800\,\,\,{\text{equiprobables}}\,\,\,\left[ {\left( {{J_m},{S_n}} \right)\,\,,\,\,\,{\text{where}}\,\,1 \leqslant m \leqslant 1000\,\,{\text{and}}\,\,\,1 \leqslant n \leqslant 800} \right]\)
\({\text{Favorable}}\,\,:\,\,60\,\,\,\,\,\left[ {\left( {{J_k},{S_k}} \right)\,\,,\,\,\,{\text{where}}\,\,1 \leqslant k \leqslant 60} \right]\,\,\,\,\)
\(? = \frac{{60}}{{1000 \cdot 800}} = \underleftrightarrow {\frac{{4 \cdot 15}}{{1000 \cdot 4 \cdot 200}}} = \frac{{4 \cdot 3 \cdot 5}}{{1000 \cdot 4 \cdot 5 \cdot 40}} = \frac{3}{{40 \cdot 1000}}\)
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
25 Aug 2021, 01:55
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Hi chetan2u,
I'm getting a bit confused here.
So in this question we can pick 60 from any one of the 2 groups, so 60/(800) * 1/1000 so we get the correct answer.
But we don't divide this by 2 factorial which I understand because either we pick senior first junior next or vice versa and then add both and then divide it by 2 because its the same.
Another question, : There are 4 distinct pair of siblings, what are the possible number of cases to form a committee of 4 with no siblings:
Here I got the answer by using 1- (exactly one sibling + exactly 2 sibling ), but that was a bit lengthy, ans = 16
so another method that results in the same answer here is the foll: 8* 6 * 4 * 2/ 4!, here we divide by 4 factorial since the order doesn't matter and we get the correct answer 16.
I understand that we divide by 4! because we have essentially formed a permutation and since order doesn't matter the number of cases reduce,
but if that's the case, then why arent we dividing by 2! here ( in the above question)
