Question

Three segments are chosen at random from six segments having lengths of 2, 3, 5, 6, 7 and 10 units. What is the probability that the three segments chosen could form a triangle? Express your answer as a common fraction

Answer 1

Using the probability concept, it is found that there is a $\frac{7}{20}$ probability that the three segments chosen could form a triangle.

A probability is the number of desired outcomes divided by the number of total outcomes.

• In this problem, the order in which the sides of the segments are chosen is not important, hence, the combination formula is used to find the number of total outcomes.

Combination formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by:

$C_{n,x} = \frac{n!}{x!(n-x)!}$

Total outcomes:

3 segments from a set of 6 lengths, hence:

$C_{6,3} = \frac{6!}{3!3!} = 20$

Desired outcomes:

• Segments can make a triangle if the sum of the lengths of the smaller segments is greater than the length of the larger segment.

• Hence, the possible options are: (2,5,6), (3,5,6), (3,5,7), (5,6,7), (5,6,10), (5,7,10), (6,7,10).

• There are 7 desired outcome, hence $D = 7$

Then:

$p = \frac{D}{T} = \frac{7}{20}$

$\frac{7}{20}$ probability that the three segments chosen could form a triangle.

To learn more about the probability concept, you can take a look at brainly.com/question/24437717