Question 1197863
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10 total, 4 defective
10-4 = 6 not defective


A = 6/10 = probability first selection is not defective
B = 5/9 = probability 2nd selection is not defective, given event A occurred prior (no replacement)
C = 4/8 = probability 3rd selection is not defective, given events A & B occurred prior (no replacement)
D = 3/7 = probability 4th selection is not defective, given events A,B,C occurred prior (no replacement)


E = probability selecting four non-defective units
E = A*B*C*D
E = (6/10)*(5/9)*(4/8)*(3/7)
E = (6*5*4*3)/(10*9*8*7)
E = (2*3*5*2*2*3)/(2*5*3*3*2*2*2*7)
E = (1)/(2*7)
E = 1/14


Answer: 1/14
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