Question

A rod of length 10 cms is broken into three parts, so that each part is having a length as an integral multiple of 1 cm. The probability that the parts are forming a triangle is:

• A. ¼
• B. ½
• C. 3/7
• D. 1/3

Answer 1

Answer: (A)

¼

Answer 2

Let x, y, z be the parts, and x ≤ y ≤ z.

Then (x, y, z) ∈ {(1, 1, 8), (1, 2, 7, (1, 3, 6), (1, 4, 5), (2, 2, 6), (2, 3, 5), (3, 3, 4), (2, 4, 4)}. Only the case when (x, y, z) ∈ {(3, 3, 4), (2, 4, 4)] we are able to form a triangle.

So, required probability will be $\frac { 2 } { 8 } = \frac { 1 } { 4 }$.