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A certain box has 10 cards written integers from 1 to 10 inc

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by Max@Math Revolution » Tue Mar 08, 2016 4:07 pm
A certain box has 10 cards written integers from 1 to 10 inclusive and the numbers written are different. If 2 different cards are selected at random, what is the probability that the sum of numbers written on the 2 cards selected less than the average (arithmetic mean) of total 10 numbers written on the 10 cards?

A. 2/45
B. 1/15
C. 4/45
D. 1/9
E. 2/15


* A solution will be posted in two days.
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by regor60 » Thu Mar 10, 2016 5:52 am
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by regor60 » Thu Mar 10, 2016 5:53 am
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by regor60 » Thu Mar 10, 2016 5:54 am
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by Max@Math Revolution » Sat Mar 12, 2016 5:24 am
A certain box has 10 cards written integers from 1 to 10 inclusive and the numbers written are different. If 2 different cards are selected at random, what is the probability that the sum of numbers written on the 2 cards selected less than the average (arithmetic mean) of total 10 numbers written on the 10 cards?

A. 2/45
B. 1/15
C. 4/45
D. 1/9
E. 2/15


-> Average in total=(1+2+...+9+10)/10=5.5. The smaller pairs than 5.5 is (1,2),(1,3),(1,4),(2,0), which makes 4 cases. Thus, probability=4C1/10C2=4/45. The answer is C.
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by Shazi1711 » Mon Mar 14, 2016 7:03 am
Max@Math Revolution wrote:A certain box has 10 cards written integers from 1 to 10 inclusive and the numbers written are different. If 2 different cards are selected at random, what is the probability that the sum of numbers written on the 2 cards selected less than the average (arithmetic mean) of total 10 numbers written on the 10 cards?

A. 2/45
B. 1/15
C. 4/45
D. 1/9
E. 2/15


-> Average in total=(1+2+...+9+10)/10=5.5. The smaller pairs than 5.5 is (1,2),(1,3),(1,4),(2,0), which makes 4 cases. Thus, probability=4C1/10C2=4/45. The answer is C.
The is an error in your solution. You cannot consider '0' since the numbers are from 1 to 10. Instead of 2,0 you should have 2,3
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