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Working through an example question in Applied Probability for Engineers and Scientists, 1st Ed., by Ephraim Suhir.

Example 1.9, beginning on p. 5, reads as follows:

A heavy, nondeformable beam is fastened to a ceiling by means of two parallel brittle rods. The weight $G$ of the beam is distributed evenly between the rods if none of them failed, and is transmitted completely to one of the rods if the other one failed. Determine the probability of non-failure and the probability of failure of the entire structure.

Define: $P_{1/2}$ as the probability of a rod failing under a load of $G/2$ and $P_1$ as the probability of a rod failing under a load of $G$. *The assumption is also made that the two rods are identical.

I had no problem obtaining the first requested probability:

$$P(\text{non-failure})=P(\text{neither fails})+P(\text{one rod fails, other rod can hold load of G})$$

$$P(\text{non-failure})=(1-P_{1/2})(1-P_{1/2})+2!P_{1/2}(1-P_1)$$

$$P(\text{non-failure})=1+P_{1/2}^2-2P_{1/2}P_1$$

For the second requested probability, the quickest way to obtain it would be to subtract the first obtained probability from 1:

$$P(\text{failure of entire structure})=1-P(\text{non-failure})$$

$$P(\text{failure of entire structure})=1-(1+P_{1/2}^2-2P_{1/2}P_1)$$

$$P(\text{failure of entire structure})=2P_{1/2}P_1-P_{1/2}^2$$

The above is the way the author does it, and he leaves it at that. However, just as an exercise, I wanted to attempt to deduce and confirm the probability of failure of the entire structure from the premises of the problem, similar to the way I computed the first requested probability. My attempt went as follows:

$$P(\text{failure of entire structure})=P(\text{both rods fail simultaneously})+P(\text{one rod fails under load of G/2, other fails under load of G})$$

$$P(\text{failure of entire structure})=P_{1/2}P_{1/2}+2!P_{1/2}P_1$$

$$P(\text{failure of entire structure})=P_{1/2}^2+2P_{1/2}P_1$$

Clearly, this result is not equivalent to the one obtained earlier. I have tried to figure out why the two methods give different results to no avail. My guess is that it has something to do with the fact that $P(\text{both rods fail simultaneously})$ requires them to fail at exactly the same instant, which would be different from two independent events that can occur one after the other.

Any enlightenment would be much appreciated.

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  • $\begingroup$ There is something puzzling about the first solution. We have non-failure if neither beam fails, or one fails and the other holds. The probability exactly one fails is $2P_{1/2}(1-P_{1/2})$, and not $2P_{1/2}$. If $P_{1/2}$ is small, the $2P_{1/2}$ gives a reasonable first-order approximation. $\endgroup$ – André Nicolas Jul 4 '15 at 22:43
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It's not quite clear how you're modelling the behaviour of the rod that holds under $G/2$ and is then supposed to carry the entire load. You didn't include the factor of $1-P_{1/2}$ for the rod holding under $G/2$ in the cases where the other one fails, but you did where the other one holds. I think the most lucid way of clearing up this confusion would be to add factors of $1-P_{1/2}$ also when the other one fails. That would make the probabilities add up to $1$, and it would allow $P_{1/2}$ to be consistently interpreted as the probability of the rod failing under $G/2$, but $P_1$ would then have to be interpreted as the conditional probability of a rod failing under $G$ given that it held under $G/2$, which is not what you introduced it as.

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To me both proposed results seem wrong. With $P_{1/2}=:p'$, $P_1=:p\>$ I argue as follows: The probability that neither of the two rods fails is $(1-p')^2$. Therefore with probability $2p'-p'^2$ we shall see the failure of a first rod, and then the second rod will fail as well with probability $p$. The probability $F$ that the beam will fall to the floor some day is therefore given by $$F=p(2p'-p'^2)\ ,$$ and the probability $N$ of this not happening is $1-F$.

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