Question

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PROBLEM 3 (30 points) A conmittee of 25-twenty women and five men -is to be seated (a) in a line (with 25 seats) at random. What is the expected number of gender-mixed adjacent pairs? (Calculate your answer up to four decimal places (rounding is OK).) answer (b) at a circular table (with 25 seats). What is the expected number of gender-mixed adjacent pairs? (Calculate your answer up to four decimal places (rounding is OK).) answer

Answer 1

a) Since there are 25 women and 5 men in a line . so for  every women you can seat a man next to her.

Number of persons = 25 women + 5 men = 30 persons

Therefore you will left with 20 single women after pairing

This is the most number of pairs that you can have

Therefore now you can seat all men together and then all women together

In this case you will have 20 pairs

Number of pairs between 20 and 5

Number of persons = 25 women + 5 men

b) For the circular table , there are 30 ways for a couple to exist

So number of seats = 30 then index them from 1 to 30 , now a couple can be found at (1,2)(2,3)........(30,1)

i.e 30 distinct places to observe a couple

After ( 30,1) we get back to where we started from i.e (1 ,2)

Therefore P( 1 , 2) = P(M1,F2) + P(M2,F1)

= 5/30 * 25/29 + 25 / 30 * 5/29

  where M1,F2 means Male at 1 & Female at 2

Similarly for (2,3), P(2,3) = P(M2,F3) + P(M3,F2) = 5/30 * 25/29 + 25 / 30 * 5/29

Total number of couples, denoted as X = X(1,2)+X(2,3)+...X(30,1)
Expected total couple E(X) = E(X(1,2)+X(2,3)+X(3,4)+....X(30,1))

Therefore E (X) = P(1,2) + P(2,1)+ ... +P(30,1)

= 5/30 * 25/29 + 25 / 30 * 5/29

= 0.1376 + 0.1411

= 0.2787

Therefore expected number is 0.2787