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Q. Out of 100 students two sections of 40 and 60 students are to be formed. If Jack and Jenny are among the 100 students, what is the probability that: a) they end up in the same section? b) they end up in different sections?

I calculated the answer of the first part just fine, it came out to be: (⁴⁰C₂+⁶⁰C₂)/¹⁰⁰C₂ =17/33

Now, to find the answer to the second part I could subtract the answer of the first part from 1 and get 16/33 which is the answer printed in my exercise book. However I’m in a bit of a dilemma.

If instead of thinking that way I say that there are two possibilities:

  1. Jack ends up in A and Jenny in B. This would give the answer as: (⁴⁰C₁.⁶⁰C₁)/¹⁰⁰C₂
  2. Jack ends up in B and Jenny in A. Which would again give the answer as: (⁴⁰C₁.⁶⁰C₁)/¹⁰⁰C₂

so the final answer would be 2(⁴⁰C₁.⁶⁰C₁)/¹⁰⁰C₂ =32/33 and this answer does not match the answer in my book.

Can somebody tell me what I’m doing wrong?

Aura Sartori
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    The probability that Jack is in $A$ and Jenny is in $B$ is half of what you say. Your denominator should be $100\times 99$ as you need to pick a slot for Jack and then a slot for Jenny. – lulu Feb 24 '18 at 16:32
  • Thanks! I got it. – Aura Sartori Feb 25 '18 at 04:40

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