The strong interactions depend only on I, not on I_z.

Preliminary: (iso)spin sums
Table of Clebsch-Gordan coefficients
For each state of two particles characterized by the (iso)spins I1 and I2:

• look at the table for I1,I2
• find the row corresponding to I_z1,I_z2
• the columns are the possible (I,I_z) states

Examples:

(a) π⁺p -> π⁺p
Isospins: I1=1, I2=1/2
Third components: I_z1=+1, I_z2=+1/2
The only possibility for (I,I_z) is (3/2,+3/2)

(b) π⁰p -> π⁰p
Isospins: I1=1, I2=1/2
Third components: I_z1=0, I_z2=+1/2
(I,I_z) can be (3/2,+1/2), coeff.=√2/3, and (1/2,+1/2), coeff.=-√1/3
Since initial and final states are equal, the amplitudes to go through (3/2,+1/2) and (1/2,+1/2) are simply 2/3 and 1/3.

(c) π^-p -> π^-p
Isospins: I1=1, I2=1/2
Third components: I_z1=-1, I_z2=+1/2
(I,I_z) can be (3/2,-1/2), coeff.=√1/3, and (1/2,-1/2), coeff.=-√2/3
Since initial and final states are equal, the amplitudes to go through (3/2,+1/2) and (1/2,+1/2) are simply 1/3 and 2/3.

(d) π^-p -> π⁰n
Isospins: I1=1, I2=1/2
Third components, initial state: I_z1=-1, I_z2=+1/2
(I,I_z) can be (3/2,-1/2), coeff.=√1/3, and (1/2,-1/2), coeff.=-√2/3
Third components, final state: I_z1=0, I_z2=-1/2
(I,I_z) can be (3/2,-1/2), coeff.=√2/3, and (1/2,-1/2), coeff.=√1/3
So, for this reaction, the amplitude to go through (3/2,-1/2) is √2/9, and to go through (1/2,-1/2) is -√2/9.

When the process proceeds through a resonance, this filters only the isospin of the resonance:
when (a),(b),(c),(d) are at the energy of the Δ (I=3/2), σ(b)/σ(a)=4/9, σ(c)/σ(a)=1/9, σ(d)/σ(a)=2/9.
(Recall that for the cross section you have to take the square of the amplitude.)
If the resonance has I=1/2, reaction (a) is forbidden.

When isospin conservation is violated, like in the EM and weak interactions, all the isospin states (compatible with the other symmetries of the system) will be equally populated. We can count the number of isospin states to infer (at least approximately) the ratios:
BR(K⁰->π⁺π^-) ~ 2*BR(K⁰->π⁰π⁰)
In fact π⁺π^- has I_z1=+1 and I_z2=-1, and so can correspond to I=0 and I=2, while π⁰π⁰ has I_z1=0 and I_z2=0, so only I=0 is possible.

Intuitive explaination: equal production rate for π⁺, π^- and π⁰. In fact, the π⁺π^- final state is the sum of "pion1 = + and pion2 = -" and viceversa.