This is an example from Hogg & Tanis' mathematical statistics textbook, 9th edition.
An electronic device has two components, $C_1$ and $C_2$, but it will operate if at least one of the components is working properly. Each component has its own switch, and both switches must be turned on, one after the other, for the device to begin operating. Thus, the device can begin to operate by either switching $C_1$ on first and then $C_2$, or vice versa. If $C_1$ is switched on first, it fails immediately with probability 0.01, whereas if $C_2$ is switched on first, it fails immediately with probability 0.02. Furthermore, if $C_1$ is switched on first and fails, the probability that $C_2$ fails immediately when it is switched on is 0.025, due to added strain. Similarly, if $C_2$ is switched on first and fails, the probability that $C_1$ fails immediately when it is switched on is 0.015. Thus, the probability that the device fails to operate after switching on $C_1$ first and then $C_2$ is $$\mathbf{P}(C_1 \text{ fails})\mathbf{P}(C_2 \text{ fails} \mid C_1 \text{ fails}) = (0.01)(0.025) = 0.00025,$$ while the probability that the device fails to operate after switching on $C_2$ first and then $C_1$ is $$\mathbf{P}(C_2 \text{ fails})\mathbf{P}(C_1 \text{ fails} \mid C_2 \text{ fails}) = (0.02)(0.015) = 0.0003.$$ The device therefore is more likely to operate properly if $C_1$ is switched on first.
From definition of conditional probability, we must have $$\mathbf{P}(C_1 \text{ fails})\mathbf{P}(C_2 \text{ fails} \mid C_1 \text{ fails}) = \mathbf{P}(C_1 \text{ fails} \cap C_2 \text{ fails}) = \mathbf{P}(C_2 \text{ fails})\mathbf{P}(C_1 \text{ fails} \mid C_2 \text{ fails}).$$ However, it's clearly not the case in this example. What is the problem here?
The thing here is that the order of the events matters, so $(A,B)$ is not the same event as $(B,A)$.