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Find the value of k for which k(24)𝑥( 42​ ) x ( x = 0, 1, 2, ...) is a pmf.(Enter the answer correct up to 2 decimal places)

Question

Find the value of k for which k(24)𝑥(42)x k(24) \cdot 𝑥(42) \cdot x ( x=0,1,2,... x = 0, 1, 2, ... ) is a pmf.\n\n(Enter the answer correct up to 2 decimal places)

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Solution

The given function is k(24)x^(42) where x = 0, 1, 2, ...

For this function to be a probability mass function (pmf), the sum of all probabilities for all possible values of x must equal 1. This is because the total probability of all outcomes in a probability space must be 1.

The sum of probabilities is given by the sum of the function for all x, which is the sum of k(24)x^(42) for x = 0, 1, 2, ...

This sum is a geometric series with first term a = k(24) * 0^(42) = 0 and common ratio r = 1.

The sum S of a geometric series with first term a, common ratio r and n terms is given by the formula S = a * (1 - r^n) / (1 - r).

Since the sum of all probabilities must be 1, we have 1 = 0 * (1 - 1^n) / (1 - 1).

Solving this equation for k gives k = 1 / 24.

Therefore, the value of k for which k(24)x^(42) is a pmf is k = 1 / 24 = 0.04 (correct to 2 decimal places).

This problem has been solved

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