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Three numbers are to be selected at random without replacement from the set of numbers {1,2,3,4..n}.

The conditional probability that the third number lies between the first two,if the first number is known to be smaller than the second is:

options:

1/3

5/6

2/3

7/12

My Approach:

I only got this

Total no of ways the number can be chosen is nC3*3!=nP3=total outcomes

A: event that the first No is smaller than second

B: event that the third No lies between first two.

p(B/A)=p(A intersection B)/p(A)

Now,I am unable to determine the the above quantities.

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There are $6$ equally-probable orders:

  • $a_1<a_2<a_3$
  • $a_1<a_3<a_2$
  • $a_2<a_1<a_3$
  • $a_2<a_3<a_1$
  • $a_3<a_1<a_2$
  • $a_3<a_2<a_1$

In $3$ of them, the first number is smaller than the second number:

  • $a_1<a_2<a_3$
  • $a_1<a_3<a_2$
  • $a_3<a_1<a_2$

In $1$ of them, the third number lies between the first number and the the second number:

  • $a_1<a_3<a_2$

Hence the probability is $\dfrac13$.

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