Coherent (or pure) state
Consider 2 basic states $\lvert0\rangle$ and $\lvert1\rangle$. (If you never heard about states, treat them as ordinary complex vectors.) Here we suppose that $\lvert0\rangle$ and $\lvert1\rangle$ are orthogonal ($\langle 0\lvert1\rangle =0$).
Now, consider $\lvert c\rangle = \frac{1}{\sqrt{2}} (\lvert0\rangle + e ^{i\phi}\lvert1\rangle)$.
$\lvert c\rangle$ is a (normalized) coherent state, because the phase $\phi$ is constant.
We may look at the density matrix, defined as $\rho = \lvert c\rangle \langle c\lvert$ (that is $\langle i\lvert\rho\lvert j\rangle=\rho_{ij} = c_i^* c_j$, with $c_1 = \frac{1}{\sqrt{2}}, c_2 = \frac{1}{\sqrt{2}}e ^{i\phi}$). We have:
$$\rho =\frac{1}{2} \begin{pmatrix} 1&e ^{i\phi}\\e ^{-i\phi}&1 \end{pmatrix}. \tag{1}$$
This density matrix describes a coherent state. You may verify that $\rho^2 = \rho$, that is $\rho$ is a projector onto the coherent state $\lvert c\rangle$.
Now, suppose the phase $\phi$ is random (that is: the phase difference between $c_1$ and $c_2$ is random), so the mean expectation value of $e ^{i\phi}$ is just zero, and we have a density matrix:
$$\rho' =\frac{1}{2} \begin{pmatrix} 1&0\\0&1 \end{pmatrix}. \tag{2}$$
The density matrix $\rho'$ does not represent no more a coherent state, this is simply a classical statistical probability law. The off-diagonal elements of the matrix $\rho$ have disappeared.
In the 2 cases, we are dealing with only one particle, and the probabilities to find the particle in the $\lvert0\rangle$ state or the $\lvert1\rangle$ state, are the same, and are $\frac{1}{2}$.
$$\langle 0\lvert\rho\lvert0\rangle= \langle 1\lvert\rho\lvert1\rangle= \langle 0\lvert\rho'\lvert0\rangle= \langle 1\lvert\rho'\lvert1\rangle = \frac{1}{2}. \tag{3}$$
Entanglement
Entanglement is about at least $2$ particles, for instance the pure state
$$ \frac{1}{\sqrt{2}}(\lvert0\rangle\lvert0\rangle+\lvert1\rangle\lvert1\rangle )\tag{4}$$is a maximally 2-particles entangled state.
From a given entanged state, you may calculate the correlations for the joint measurement of the 2 particles.
It appears that these entangled quantum correlations are stronger than the classical statistical correlations.
Correlations do not mean that you may exchange instantaneous information; see also this previous answer.