To solve for \(P(X = 0)\) in the given binomial distribution, we need to use the properties of a binomial distribution and the information provided about its mean and standard deviation.

The parameters of a binomial distribution are:

• \(n\): Number of trials
• \(p\): Probability of success on a single trial

The mean \(\mu\) of a binomial distribution is given by:

\(\mu = n \cdot p\)

The standard deviation \(\sigma\) of a binomial distribution is given by:

\(\sigma = \sqrt{n \cdot p \cdot (1-p)}\)

From the problem, we know:

• The mean \(\mu = 6\)
• The standard deviation \(\sigma = \sqrt{2}\)

We set up the equations using the given mean and standard deviation:

\(n \cdot p = 6\) (Equation 1)

\(\sqrt{n \cdot p \cdot (1-p)} = \sqrt{2}\)

Squaring both sides of the second equation gives:

\(n \cdot p \cdot (1-p) = 2\) (Equation 2)

Now, substituting the value of \(n \cdot p\) from Equation 1 into Equation 2:

\(6 \cdot (1-p) = 2\)

\(6 - 6p = 2\)

Solving for \(p\):

\(4 = 6p\)

\(p = \frac{2}{3}\)

Substituting \(p = \frac{2}{3}\) back into Equation 1 to solve for \(n\):

\(n \cdot \frac{2}{3} = 6\)

\(n = 6 \cdot \frac{3}{2} = 9\)

The probability \(P(X = 0)\) is given by the binomial probability formula for \(X = 0\):

\(P(X = 0) = \binom{n}{0} \cdot p^0 \cdot (1-p)^n\)

Where:

• \(n = 9\)
• \((1-p) = \frac{1}{3}\)

Therefore,

\(P(X = 0) = (1-p)^n = \left(\frac{1}{3}\right)^9\)

Hence, the value of \(P(X = 0)\) is \(\left(\frac{1}{3}\right)^9\).