Five line segments are given, with lengths 1 , 3 , 5 , 7 and…
Five line segments are given, with lengths $1$, $3$, $5$, $7$ and $9$. Find the probability that three randomly chosen segments can form a triangle.
Explanation of Method
Three segments form a triangle exactly when the two shorter ones together are longer than the longest one (the triangle inequality). There are only $\binom53=10$ equally likely ways to choose three segments, so we list them all, check the inequality for each, and count.
Step 1: The triangle condition
Three segments of lengths $a\le b\le c$ form a triangle if and only if
\[ a + b > c. \]
(The other two inequalities, $a+c>b$ and $b+c>a$, hold automatically because $c$ is the longest.) If $a+b\le c$, the two short sides cannot reach each other when laid along the long one.
Step 2: All outcomes
Choosing three segments out of five, without order:
\[ \binom{5}{3} = 10 \]
equally likely choices.
Step 3: Check every triple
| Triple | Sum of the two shorter | Longest | Triangle? |
|---|---|---|---|
| $\{1,3,5\}$ | $1+3=4$ | $5$ | no |
| $\{1,3,7\}$ | $1+3=4$ | $7$ | no |
| $\{1,3,9\}$ | $1+3=4$ | $9$ | no |
| $\{1,5,7\}$ | $1+5=6$ | $7$ | no |
| $\{1,5,9\}$ | $1+5=6$ | $9$ | no |
| $\{1,7,9\}$ | $1+7=8$ | $9$ | no |
| $\{3,5,7\}$ | $3+5=8$ | $7$ | yes |
| $\{3,5,9\}$ | $3+5=8$ | $9$ | no |
| $\{3,7,9\}$ | $3+7=10$ | $9$ | yes |
| $\{5,7,9\}$ | $5+7=12$ | $9$ | yes |
Any triple containing the segment of length $1$ fails: the other two differ by at least $2$, so $1 + b \le c$.
Step 4: The probability
Favourable triples: $\{3,5,7\}$, $\{3,7,9\}$, $\{5,7,9\}$ — three of ten.
\[ \boxed{P = \frac{3}{10} = 0.3.} \]
Result
| Answer | |
|---|---|
| All choices | $\binom53=10$ |
| Triangles | $\{3,5,7\},\ \{3,7,9\},\ \{5,7,9\}$ |
| Probability | $P=3/10=0.3$ |