what is the probability that third number lies between first two if the first number is known to be smaller than the second?

Question

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.

Answer 1

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 $\dfrac13$.