Question

A personnel director interviewing 12 senior engineers for five job openings has scheduled seven interviews for the first day and five for the second day of interviewing. Assume that the candidates are interviewed in a random order.
(a) What is the probability that x of the top four candidates are interviewed on the first day?
h(N; 5, 5, 12)
h(x; 5, 12, 5)
h(N; 7, 12, 5)
h(x; 7, 5, 12)
(b) How many of the top four candidates can be expected to be interviewed on the first day? (Round your answer to two decimal places.)

Answer 1

Answer:

a) h(x; 7, 5, 12) = (⁵Cₓ)( ⁷C₇₋ₓ) / (¹²C₇)

b) 2.92

Step-by-step explanation:

a)

Here

Number of interviewees = N = 12

Number of job openings = M = 5

Interviews schedules for the first day =  n = 7

N − M = 12 - 5 = 7

Using hypergeometric distribution:

Let X be the no. top four candidates interviewed on first day.

The probability mass function of X:

P(X = x) = $(^{M} C_{x})$   $(^{N-M} C_{n-x})$ /  $(^{N} C_{n})$

It can be written as:

h(x; n, M, N)  =  $(^{M} C_{x})$   $(^{N-M} C_{n-x})$ /  $(^{N} C_{n})$    

                    = (5Cx) (7C7-x) / (12C7)

                    = (⁵Cₓ)( ⁷C₇₋ₓ) / (¹²C₇)

h(x; 7, 5, 12) = (⁵Cₓ)( ⁷C₇₋ₓ) / (¹²C₇)

b)

The expectation is: E(X) = np

E(X) = n * M/N

              = 7 * 5/12

              = 7 * 0.41667

              = 2.9167