91Ó°ÊÓ

In probability, the sample space represents all the possible outcomes of a certain experiment. For the problem at hand, we have three concrete beams that can fail in two distinct ways: by shear \((S)\) or flexure \((F)\). Understanding this leads us to explore all combinations of these outcomes for the experiment of testing three random beams.When constructing a sample space, list every conceivable arrangement of the outcomes. For our exercise, each of the three beams could fail by either \(S\) or \(F\). The sample space is:

• SSS
• SSF
• SFS
• SFF
• FSS
• FSF
• FFS
• FFF

This sample space highlights the principle of calculating total possibilities by using tree diagrams or permutation reasoning, showing all potential sequences of outcomes.

A random variable is a function that assigns numerical values to each outcome in a sample space. It's a critical component in probability and statistics because it translates complex outcomes into manageable numbers. In the context of our exercise, we define the random variable \(X\). This variable stands for the number of beams out of the three that failed by shear.By examining each outcome in our sample space:

• For \(SSS\), all three beams fail by shear, so \(X = 3\).
• For \(SSF\), two beams fail by shear, so \(X = 2\).
• For \(SFS\), two beams fail by shear, so \(X = 2\).
• For \(SFF\), one beam fails by shear, so \(X = 1\).
• For \(FSS\), two beams fail by shear, so \(X = 2\).
• For \(FSF\), one beam fails by shear, so \(X = 1\).
• For \(FFS\), one beam fails by shear, so \(X = 1\).
• For \(FFF\), no beams fail by shear, so \(X = 0\).

This systematic approach allows us to deduce meaningful information about the behavior of the beams based on experimental variance.

Combinatorics is the branch of mathematics dealing with counting and organizing possibilities. In probability, it helps determine the possible outcomes and probabilities. Here, the combinatorial aspect lies in counting outcomes of shear or flexure failures.With two failure types (shear \((S)\) and flexure \((F)\)) for each beam, the number of total outcomes for three beams is computed by multiplying the number of choices for each beam:

• First beam: 2 choices (\(S\) or \(F\))
• Second beam: 2 choices
• Third beam: 2 choices

The total combinations are \(2 \times 2 \times 2 = 8\), matching the sample space size.Through combinatorics, one can also delve into probability calculation when questions require. Knowing how to arrange or count these outcomes is a base skill in designing and assessing probability models.