Question

Five men and 5 women are ranked according to their scores on an examination. Assume that no two scores are alike and all possible rankings are equally likely. Let X denote the highest ranking achieved by a woman(for instance X=1 if the top ranked person is female) Find P(X=i), I=1,2,3,4,,,,8,9,10. Response: Let Ei be the event that the the ith scorer is female. Then the event X = i correspdonds to the event Ec1Ec2 · · ·Ei. It follows that P{X = i } = P(Ec1Ec2 · · ·Ei)= P(Ec1)P(Ec2|c1) · · · P(Ei|Ec1 · · ·Eci−1). i | P{x=i} 1 | (1/2) 2 | (5/18 3 | (5/36) 4 | (5/84) 5 | (5/252) 6 | (1/252) 7,8,9,10 | 0. How could it be 1 is (1/2) and 2 is (5/18) and so on.... Thank you...

Answer 1

Answer:

P(X=i) where i = 1,2,3,4,5,6,7,8,9,10 = 1/2, 5/18, 5/36, 5/84, 5/252, 1/252, 0, 0, 0, 0.

Step-by-step explanation:

X denotes the highest ranking achieved by the woman.

When X=1, the top ranked person is a female.

When X=2, the first person is a male and the second ranked person is a female.

Similarly, when X=3, the first two ranked persons are male and the third one is a female.

When X=4, the first three persons are male and the fourth one is a female.

When X=5, the first four persons are males and the fifth person is a female.

When X=6, the first five people are males and the sixth person is a female. The rest of the four people are also females since there are only five men in a sample space.

The probability for X=7, 8, 9, 10 is zero because there are only five men who can achieve the first five positions and the last highest rank that can be achieved by a woman is 6.

To compute the probabilities, we will use the formula:

No. of ways a female can be ranked X/Total number of ways to rank 10 people

Note that the total number of ways of ranking 10 different people is 10P10 or 10!

For X=1, the first position can be taken by any of the 5 women. The possible ways of the first person being a woman is 5C1. The rest of the 9 people can take any of the ranks. They can be ordered in 9P9 ways.

So, P(X=1) = (5C1)(9P9)/(10P10) = (5 x 362880)/(3628800) = 1/2

For X=2, the first rank must be taken by a male. The number of ways to arrange the first person as a male out of the 5 men can be calculated by 5P1. The second position must be taken by a female and rest of the 8 positions can be taken by any of the 8 people in 8P8 ways.

So, P(X=2) = (5P1)(5C1)(8P8)/(10P10) = (5 x 5 x 40320)/(3628800) = 5/18

For X=3, first two people must be men and the number of ways to arrange 2 out of 5 males at the first two positions is 5P2. The third position is a female. The rest of the 7 people can be ordered in 7P7 ways.

P(X=3) = (5P2)(5C1)(7P7)/(10P10) = (20 x 5 x 5040)/(3628800) = 5/36

P(X=4) = (5P3)(5C1)(6P6)/(10P10) = (60 x 5 x 720)/(3628800) = 5/84

P(X=5) = (5P4)(5C1)(5P5)/(10P10) = (120 x 5 x 120)/(3628800) = 5/252

P(X=6) = (5P5)(5C1)(4P4)/(10P10) = (120 x 5 x 24)/(3628800) = 1/252

P(X=7) = 0

P(X=8) = 0

P(X=9) = 0

P(X=10) = 0