Seven sticks with lengths 2, 3, 5, 7, 11, 13 and 17 inches are placed in a box. Three of the sticks are randomly selected. What is the probability that a triangle can be formed by joining the endpoints of the sticks? Express your answer as a common fraction.
By the triangle inequality, three segments form a triangle if and only if the sum of the smaller two lengths exceeds the greatest length.  Therefore, if $2$ is one of the sticks drawn, then the three sticks cannot be used to form a triangle.  If 3 is the smallest length drawn, then the possible sets of sticks are (3,5,7) and (3,11,13).  If 5 is the smallest length drawn, then (5,7,11), (5,11,13), and (5,13,17) are the sets that satisfy the triangle inequality.  If 7 is the smallest length drawn, then (7,11,13), (7,11,17), (7,13,17) all satisfy the triangle inequality.  Finally, (11,13,17) satisfies the triangle inequality.  In total, there are $2+3+3+1=9$ sets of sticks that could be used to form a triangle.  There are $\binom{7}{3}=35$ equally likely sets of 3 sticks, so the probability that one of the 9 sets that form a triangle will be chosen is $\boxed{\frac{9}{35}}$.
\frac{9}{35}
Level 5
Counting & Probability
Problem: Seven sticks with lengths 2, 3, 5, 7, 11, 13 and 17 inches are placed in a box. Three of the sticks are randomly selected. What is the probability that a triangle can be formed by joining the endpoints of the sticks? Express your answer as a common fraction.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine the probability that three randomly selected sticks can form a triangle, we must apply the triangle inequality theorem. This theorem states that for any three lengths a, b, and c, they can form a triangle if and only if the sum of the lengths of any two sides is greater than the length of the third side. 

First, we calculate the total number of ways to choose three sticks from the seven available. This is given by the combination formula C(n, k), where n is the total number of items to choose from, and k is the number of items to choose. Here, n = 7 and k = 3:

C(7, 3) = 7! / (3!(7-3)!) = 35.

Next, we need to identify which combinations of the selected sticks satisfy the triangle inequality. We will systematically check each combination of three sticks:

1. (2, 3, 5): 2 + 3 = 5, not greater than 5.
2. (2, 3, 7): 2 + 3 = 5, not greater than 7.
3. (2, 3, 11): 2 + 3 = 5, not greater than 11.
4. (2, 3, 13): 2 + 3 = 5, not greater than 13.
5. (2, 3, 17): 2 + 3 = 5, not greater than 17.
6. (2, 5, 7): 2 + 5 = 7, not greater than 7.
7. (2, 5, 11): 2 + 5 = 7, not greater than 11.
8. (2, 5, 13): 2 + 5 = 7, not greater than 13.
9. (2, 5, 17): 2 + 5 = 7, not greater than 17.
10. (2, 7, 11): 2 + 7 = 9, greater than 11.
11. (2, 7, 13): 2 + 7 = 9, not greater than 13.
12. (2, 7, 17): 2 + 7 = 9, not greater than 17.
13. (2, 11, 13): 2 + 11 = 13, not greater than 13.
14. (2, 11, 17): 2 + 11 = 13, not greater than 17.
15. (2, 13, 17): 2 + 13 = 15, not greater than 17.
16. (3, 5, 7): 3 + 5 = 8, greater than 7.
17. (3, 5, 11): 3 + 5 = 8, not greater than 11.
18. (3, 5, 13): 3 + 5 = 8, not greater than 13.
19. (3, 5, 17): 3 + 5 = 8, not greater than 17.
20. (3, 7, 11): 3 + 7 = 10, greater than 11.
21. (3, 7, 13): 3 + 7 = 10, not greater than 13.
22. (3, 7, 17): 3 + 7 = 10, not greater than 17.
23. (3, 11, 13): 3 + 11 = 14, greater than 13.
24. (3, 11, 17): 3 + 11 = 14, greater than 17.
25. (3, 13, 17): 3 + 13 = 16, not greater than 17.
26. (5, 7, 11): 5 + 7 = 12, greater than 11.
27. (5, 7, 13): 5 + 7 = 12, greater than 13.
28. (5, 7, 17): 5 + 7 = 12, not greater than 17.
29. (5, 11, 13): 5 + 11 = 16, greater than 13.
30. (5, 11, 17): 5 + 11 = 16, not greater than 17.
31. (5, 13, 17): 5 + 13 = 18, greater than 17.
32. (7, 11, 13): 7 + 11 = 18, greater than 13.
33. (7, 11, 17): 7 + 11 = 18, greater than 17.
34. (7, 13, 17): 7 + 13 = 20, greater than 17.
35. (11, 13, 17): 11 + 13 = 24, greater than 17.

After checking all combinations, the valid combinations that satisfy the triangle inequality are:
- (3, 5, 7)
- (3, 7, 11)
- (3, 11, 13)
- (5, 7, 11)
- (5, 7, 13)
- (5, 11, 13)
- (5, 13, 17)
- (7, 11, 13)
- (7, 11, 17)
- (7, 13, 17)
- (11, 13, 17)

This gives us a total of 9 valid combinations. 

To find the probability, we divide the number of successful outcomes by the total number of outcomes:

Probability = Number of successful outcomes / Total outcomes = 9 / 35.

Thus, the final answer is $\frac{9}{35}$.<|END_OF_LATENT|> The final answer is $\frac{9}{35}$.