# Why does the probability of obtaining a value of a measurement follow from Dirac's general assumption?

In Dirac's The Principle of Quantum Mechanics he makes the general assumption that "if the measurement of the observable $\xi$ for the system in the state corresponding to $|x\rangle$ is made a large number of times, the average of all the results obtained will be $\langle x|\xi|x\rangle$, provided $|x\rangle$ is normalized."

Then he states that the average value of any function of $\xi$, $f(\xi)$ is $\langle x|f(\xi)|x\rangle$. Now, taking $f(\xi)$ to be the function that is equal to unity when $\xi=a$ and to be zero otherwise, noted by $\delta_{\xi a}$, the average value of this function is the probability $P_a$ of obtaining the result $a$ when we perform a single measurement of the observable $\xi$.

Now I know that this really gives the probability, because $\langle x|\delta_{\xi a}|x\rangle=|c_a|^2$, where $c_a$ is the coefficient of the eigenstate of $\xi$ corresponding to the eigenvalue $a$ in the expansion of $|x\rangle$. But Dirac seems to "know" that this is the probability without "knowing" that $|c_a|^2$ is the probability of obtaining a certain result. So my question is: how do we know that this expression $\langle x|\delta_{\xi a}|x\rangle$ is really the probability $P_a$ without using the expansion coefficients?

The Projection Operator $\delta_{\varepsilon\, a}$ is $1$ for some particular $|a\rangle$ and is $0$ on any orthogonal state. The definition of the expectation in classical probability theory for this operator is given by \begin{equation} \left\langle \delta_{\varepsilon\, a}\right\rangle = 1\Pr(|x\rangle = |a\rangle) + 0\Pr(|x\rangle \perp |a\rangle) = \Pr(|x\rangle = |a\rangle)\end{equation}i.e the expectation of the projection operator is the probability that $|x\rangle$ is found in state $|a\rangle$ when a measurement is made. However we can also write \begin{align} \left\langle \delta_{\varepsilon\, a}\right\rangle &= \langle x|\delta_{\varepsilon\, a}|x\rangle\\ &= \langle x|a\rangle\langle a|x\rangle\\ & = |\langle a|x\rangle|^2\end{align} So the rule for finding probabilities in QM can be derived from the rule for finding expectations of operators.