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A certain junior class has 1,000 students and a certain

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Re: A certain junior class has 1000 students and a certain  [#permalink]

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New post 22 Feb 2018, 17:45
gurpreet07 wrote:
A certain junior class has 1000 students and a certain senior class has 800 students. Among these students, there are 60 siblings pairs each consisting of 1 junior and 1 senior. If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?

A. 3/40000
B. 1/3600
C. 9/2000
D. 1/60
E. 1/15


The probability of selecting any one sibling from the 60 sibling pairs in the junior class is 60/1000. Once that person is selected, the probability of selecting his or her sibling from the senior class is 1/800; thus, the probability of a selecting a sibling pair is:

60/1000 x 1/800 = 3/50 x 1/800 = 3/40000

Alternatively, the probability of selecting any one sibling from the 60 sibling pairs in the senior class is 60/800. Once that person is selected, the probability of selecting his or her sibling from the junior class is 1/1000; thus, the probability of a selecting a sibling pair is:

60/800 x 1/1000 = 3/40 x 1/1000 = 3/40000

Answer: A
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Re: A certain junior class has 1,000 students and a certain  [#permalink]

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New post 23 Feb 2018, 08:37
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Top Contributor
blog wrote:
A certain junior class has 1,000 students and a certain senior class has 800 students. Among these students, there are 60 siblings pairs, each consisting of 1 junior and 1 senior. If 1 student is to be selected at random from each class, what is the probability that the 2 students selected at will be a sibling pair?

A. 3/40,000
B. 1/3,600
C. 9/2,000
D. 1/60
E. 1/15


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
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Re: A certain junior class has 1,000 students and a certain  [#permalink]

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New post 16 Sep 2018, 15:07
GMATPrepNow wrote:
blog wrote:
A certain junior class has 1,000 students and a certain senior class has 800 students. Among these students, there are 60 siblings pairs, each consisting of 1 junior and 1 senior. If 1 student is to be selected at random from each class, what is the probability that the 2 students selected at will be a sibling pair?

A. 3/40,000
B. 1/3,600
C. 9/2,000
D. 1/60
E. 1/15


\(\left( J \right)\,\,{\text{Junior}}\,\,:\,\,\,1000\,\,{\text{students}}\,\,,\,\,\,{J_1}\,,\,{J_2},\,\, \ldots \,\,,\,\,{J_{60}}\,\,{\text{among}}\,\,{\text{them}}\,\,\,\left( {{\text{the}}\,\,{\text{ones}}\,\,{\text{with}}\,\,{\text{senior}}\,\,{\text{siblings}}} \right)\)

\(\left( S \right)\,\,{\text{Senior}}\,\,:\,\,\,800\,\,{\text{students}}\,\,,\,\,\,{S_1}\,,\,{S_2},\,\, \ldots \,\,,\,\,{S_{60}}\,\,{\text{among}}\,\,{\text{them}}\,\,\,\left( {{\text{the}}\,\,{\text{ones}}\,\,{\text{with}}\,\,{\text{junior}}\,\,{\text{siblings}}} \right)\)

\(\left( {{J_1}\,,\,{S_1}} \right)\,\,;\,\,\left( {{J_2}\,,\,{S_2}} \right)\,\,;\,\, \ldots \,\,;\,\,\left( {{J_{60}}\,,\,{S_{60}}} \right)\,\,\,:\,\,\,{\text{pairs}}\,\,{\text{of}}\,\,{\text{siblings}}\,\)


\(? = P\left( {{\text{pair}}\,{\text{of}}\,{\text{siblings}}\,,\,\,{\text{in}}\,{\text{one}}\,J\,{\text{and}}\,{\text{one}}\,S\,{\text{extraction}}} \right)\)


\({\text{Total}}\,\,:\,\,\,1000 \cdot 800\,\,\,{\text{equiprobables}}\,\,\,\left[ {\left( {{J_m},{S_n}} \right)\,\,,\,\,\,{\text{where}}\,\,1 \leqslant m \leqslant 1000\,\,{\text{and}}\,\,\,1 \leqslant n \leqslant 800} \right]\)

\({\text{Favorable}}\,\,:\,\,60\,\,\,\,\,\left[ {\left( {{J_k},{S_k}} \right)\,\,,\,\,\,{\text{where}}\,\,1 \leqslant k \leqslant 60} \right]\,\,\,\,\)


\(? = \frac{{60}}{{1000 \cdot 800}} = \underleftrightarrow {\frac{{4 \cdot 15}}{{1000 \cdot 4 \cdot 200}}} = \frac{{4 \cdot 3 \cdot 5}}{{1000 \cdot 4 \cdot 5 \cdot 40}} = \frac{3}{{40 \cdot 1000}}\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: Yep, I know it is easy but probabilty is still giving me  [#permalink]

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Re: Yep, I know it is easy but probabilty is still giving me   [#permalink] 08 Oct 2019, 06:08

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