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Two different numbers are chosen from the set {1, 2, 3, ..., 10}.  What is the probability that their sum is 8 or greater?

 Jul 20, 2020
 #1
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First, let's decide what the denomintor will be. In this case, it is 10 times 9, which is 90. Since there can be repeats of the same group, we devide by 2!, which is 2, and that will give us 90

 

From here, let's see what combos will end up being greater than or equal to 8 by using a table:

Number.   1.    2     3.   4.   5.   6.   7.  8.   9.   10    Total:    
100000011114
200000111115
300001111116
400001111116
500110111117
601111011118
711111101119
811111110119
911111111019
1011111111109

Now, we have to add all the totals up. This is 4+5+6(2)+7+8+4(9) = 72. Dividing by 2, we get 36. Therefore, the probability is 36/45, which is 4/5.

 Jul 20, 2020
edited by ilorty  Jul 21, 2020
 #2
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The following are number of ways you can choose 2 numbers out of 10:
{1, 2} | {1, 3} | {1, 4} | {1, 5} | {1, 6} | {1, 7} | {1, 8} | {1, 9} | {1, 10} | {2, 3} | {2, 4} | {2, 5} | {2, 6} | {2, 7} | {2, 8} | {2, 9} | {2, 10} | {3, 4} | {3, 5} | {3, 6} | {3, 7} | {3, 8} | {3, 9} | {3, 10} | {4, 5} | {4, 6} | {4, 7} | {4, 8} | {4, 9} | {4, 10} | {5, 6} | {5, 7} | {5, 8} | {5, 9} | {5, 10} | {6, 7} | {6, 8} | {6, 9} | {6, 10} | {7, 8} | {7, 9} | {7, 10} | {8, 9} | {8, 10} | {9, 10} (total: 45)

 

Their sums =(3, 4, 5, 5, 6, 6, 7, 7, 7, 8, 8, 8, 9, 9, 9, 9, 10, 10, 10, 10, 11, 11, 11, 11, 11, 12, 12, 12, 12, 13, 13, 13, 13, 14, 14, 14, 15, 15, 15, 16, 16, 17, 17, 18, 19)>Total = 45


There are 39 out of 45 which are 8 and >. Therefore, the probability is: 39 / 45

 Jul 20, 2020

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