What is the value of k?

Calculation:

Given,

Let X be a random variable following a binomial distribution with parameters n = 6 and p = k.

Further, it is given that:

\( 9P(X = 4) = P(X = 2) \)

The probability mass function for a binomial distribution is:

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

For P(X = 4) , we have:

\( P(X = 4) = \binom{6}{4} k^4 (1 - k)^2 = 15 k^4 (1 - k)^2 \)

For P(X = 2) , we have:

\( P(X = 2) = \binom{6}{2} k^2 (1 - k)^4 = 15 k^2 (1 - k)^4 \)

We are given that:

\( 9P(X = 4) = P(X = 2) \)

Substitute the expressions for P(X = 4)  and P(X = 2) :

\( 9 \times 15 k^4 (1 - k)^2 = 15 k^2 (1 - k)^4 \)

Cancel the common factor of 15:

\( 9 k^4 (1 - k)^2 = k^2 (1 - k)^4 \)

\( 9 k^2 = (1 - k)^2 \)

\( 9 k^2 = 1 - 2k + k^2 \)

\( 8 k^2 + 2k - 1 = 0 \)

Use the quadratic formula to solve for k :

\( k = \frac{-2 \pm \sqrt{2^2 - 4 \times 8 \times (-1)}}{2 \times 8} \)

\( k = \frac{-2 \pm \sqrt{4 + 32}}{16} \)

\( k = \frac{-2 \pm \sqrt{36}}{16} \)

\( k = \frac{-2 \pm 6}{16} \)

Thus, the two possible values for k are:

\( k = \frac{4}{16} = \frac{1}{4} \) or \( k = \frac{-8}{16} = -\frac{1}{2} \)

Since k  represents a probability, it must be between 0 and 1. Therefore, the valid solution is:

\( k = \frac{1}{4} \)

Hence, the correct answer is Option 3.