To make 2-qubits entanglement on quantum circuit

First

To calculate on quantum computer we think about state vector as a simulator. Now we see the 2-qubits entanglement.

The state vector of 2-qubits superposition is,

[1
1
1
1]

These 4 element of vector denote state probability of 00,01,10,11. If we can control these probability we can make variety of applications using superposition and entanglement expected for quantum speedup.

Basic of quantum entanglement

The famous quantum circuit of entanglement is,

---H---*----
|
-------X----

And the state vector is,

[1
0
0
1]

Now we have probabilty of only 00 and 11. This circuit select 50% of 00 and 11 from 4 possibility of 00,01,10 and 11.

To control state vector

For example, when we have,

[1
0
0
0]

This is the initial state of 2-qubits of 00. We can get this state vector from 2 |0> qubits.

[1 ⊗[1
0] 0]

And with X pauli gate we can change the state vector changing the shape of state vector like,

[0  or [0  or [0
1 0 0
0 1 0
0] 0] 1]

Next if we want to have just 2 probability from 4 state like,

[1
0
1
0]

We can get this state vector by applying H gate for just 1 qubit.

And if we want all probability for all 4 state vector,

[1
1
1
1]

We can get this state by applying H gate for both 2-qubits.

How to get 3 from 4

To get 3 probability from 4 state vector element like,

[1
0
1
1]

Now we realize it using RY gate. First we have,

[1
0
0
0]

By rotating the first qubits 1/3 of degree we get,

[1
0
√2
0]

And then we apply Controlled-RY gate for this state vector we can get the expected state vector. The unitary matrix of Controlled-RY gate is,

[1 
1
1/√2 -1/√2
1/√2 1/√2 ]

And we finally get,

[1
0
1
1]

as expected state vector.

Conclusion

To get 3 probability from 4 state vector, we can use RY gate and Controlled-RY gate to realize what we want.