A personnel director interviewing 11 senior engineers for four job openings has scheduled six interviews for the first day and five for the second day of interviewing. Assume that the candidates are interviewed in random order. a. What is the probability that x of the top four candidates are interviewed on the first day? b. ?How many of the top four candidates can be expected to be interviewed on the first day?

Answer: Step1: Given, a personnel director interviewing 11 senior engineers for four job openings has scheduled six interviews for the first day and five for the second day of interviewing. Assume that the candidates are interviewed in random order. Here, N = 11, M = 4, n = 6, N - M = 7. Step2: a). To find the probability that ‘x’ of the top four candidates are interviewed on the first day. X = number of top four candidates are interviewed on the first day. (x )( NM) P(X = x) = Nn x ( n) 4 114 P(X = x) = (x)( 6 x , x = 0,1,2,3,4. (6) Substitute, x = 0,1,2,3,4 in the above equation. 4 114 ( 0( 6 0 P(X = 0) = ( 61 1×7 = 462 P(X = 0) = 0.0151 4 114 (1)( 6 1 P(X = 1) = ( ) 6 4×21 = 462 P(X = 1) = 0.1818 4 114 ( 2( 6 2 P(X = 2) = ( )1 6 = 6×35 462 P(X = 2) = 0.4545 4 114 (3)( 6 3 P(X = 3) = (6) = 4×35 462 = 0.4545 4 114 ( 4( 6 4 P(X = 4) = 11 ( 6 = 1×21 462 = 0.0454 The number of top four candidates are interviewed on the first day are tabulated below x 0 1 2 3 4 p(x) 0.0151 0.1818 0.4545 0.4545 0.0454 Step3: b). To find the top four candidates can be expected to be interviewed on the first day. Mean E(x) = n.( M ) 4 = 6.(11) = 2.1818. Therefore, the top four candidates can be expected to be interviewed on the first day is 2.1818.