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14 Mar 2005, 22:12
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1. Crown throws 3 dice and records the product of the numbers appearing at the top of each die as the result of the attempt. What is the probability that the result of any attempt is an odd integer divisible by 25?
A) 7/216
B) 5/91
C) 13/88
D) 1/5
E 3/8
A) 7/216
B) 5/91
C) 13/88
D) 1/5
E 3/8
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14 Mar 2005, 22:33
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A for me : 7/216
maximum number with 3 dices : 6*6*6 = 216
maximum integer divisible by 25 : 5*5*6 = 150
odd integer divisible by 25 and <=150 : 25, 75, 125
25 -> 5*5*1, 5*1*5, 1*5*5
75 -> 5*5*3, 5*3*5, 3*5*5
125 -> 5*5*5
total : 7 possibilities
total potential outcome : 216
7/216
maximum number with 3 dices : 6*6*6 = 216
maximum integer divisible by 25 : 5*5*6 = 150
odd integer divisible by 25 and <=150 : 25, 75, 125
25 -> 5*5*1, 5*1*5, 1*5*5
75 -> 5*5*3, 5*3*5, 3*5*5
125 -> 5*5*5
total : 7 possibilities
total potential outcome : 216
7/216
