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A coaching held separate workshops for training in three languages.

50 students participated in workshops and learnt one or more languages. 13 students learnt a single language, 25 students learnt 2 languages and 12 students learnt 3 languages.

If three students are selected at random, find the probability that exactly one has attended 3 workshops, exactly one has attended two workshops and one has attended a single workshop?

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up vote 2 down vote accepted

Please check my attempt for correctness:

choosing 3 out of 50 children can be done in 50C3 ways.

And to satisfy the condition, i can make a selection of 1 from 13, 1 from 25 and 1 from 12 students.

so i have 12*13*25/(50C3).

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Which is $\frac{3900}{19600}=\frac{39}{196}\approx 0.199$. Yes. – Brian M. Scott Jul 1 '12 at 22:18
Thanks. Perhaps the answer i have with me is incorrect. It states 78/2500 (calculated by 12/50 * 13/50 * 25/50) – Karan Jul 1 '12 at 22:20
That’s based on a different interpretation of at random. You and I calculated it on the assumption that the three kids were chosen at one time. That answer is based on the assumption that one kid was chosen and put back into the pool, then a second was chosen and put back into the pool, and then a third was chosen, so that in principle you could have chosen the same kid all three times. – Brian M. Scott Jul 1 '12 at 22:27
But then how do i determine which approach to follow when i encounter such questions? There is no mention in the question that kids are selected with replacement. – Karan Jul 2 '12 at 8:55
I would have made exactly the same assumption that you did. It’s simply a badly worded question, and there’s not much that you can do about those. – Brian M. Scott Jul 2 '12 at 16:57

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