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For ∣ψ⟩=cos⁡(θ)∣0⟩+sin⁡(θ)∣1⟩|\psi\rangle = \cos(\theta)|0\rangle + \sin(\theta)|1\rangle with θ=π/6\theta = \pi/6, and ∣+⟩=(∣0⟩+∣1⟩)/2|+\rangle = (|0\rangle + |1\rangle)/\sqrt{2}, what is p(+)=∣⟨+∣ψ⟩∣2=(1/2)(cos⁡(θ)+sin⁡(θ))2p(+) = |\langle+|\psi\rangle|^2 = (1/2)(\cos(\theta) + \sin(\theta))^2?