The probability of finding a particle in a classically forbidden region in the ground state of a simple harmonic oscillator can be explained using quantum mechanics principles.

Classically Forbidden Region:
In classical mechanics, the particle in a simple harmonic oscillator can only oscillate within a certain range determined by its energy. However, in quantum mechanics, the particle can have a non-zero probability of being found in regions where its energy is classically forbidden, i.e., regions where the particle's energy is lower than the potential energy of the oscillator.

Ground State of Simple Harmonic Oscillator:
The ground state of a simple harmonic oscillator is its lowest energy state, where the particle is most likely to be found. In this state, the particle has a well-defined energy and position.

Probability Distribution:
The probability distribution of finding the particle in a certain region can be represented by the wavefunction of the system, which is given by the square of the wavefunction's magnitude, |Ψ(x)|^2. The wavefunction describes the behavior of the particle in terms of its position.

Wavefunction of Simple Harmonic Oscillator:
In the case of a simple harmonic oscillator, the wavefunction can be expressed as a Gaussian function, which oscillates back and forth around the equilibrium position. The ground state wavefunction is given by:

Ψ(x) = (mω/πħ)^0.25 * exp(-mωx^2/2ħ),

where m is the mass of the particle, ω is the angular frequency of the oscillator, x is the position of the particle, and ħ is the reduced Planck's constant.

Calculating the Probability:
To calculate the probability of finding the particle in a classically forbidden region, we need to integrate the square of the wavefunction over that region.

Let's assume the classically forbidden region is defined by the range x1 to x2. The probability is then given by:

P = ∫(x1 to x2) |Ψ(x)|^2 dx.

Substituting the expression for the ground state wavefunction, we have:

P = ∫(x1 to x2) [(mω/πħ)^0.25 * exp(-mωx^2/2ħ)]^2 dx.

Simplifying the expression, we get:

P = (mω/πħ)^0.5 * ∫(x1 to x2) exp(-mωx^2/ħ) dx.

The integral in this expression is evaluated using standard techniques, resulting in a constant factor multiplied by the error function:

P = (mω/πħ)^0.5 * (1/2) [erf((x2√(mω))/√ħ) - erf((x1√(mω))/√ħ)].

This expression gives the probability of finding the particle in the classically forbidden region. To find the specific value of 0.18, the values of x1, x2, m, ω, and ħ need to be provided.