Question

Suppose \(X\) has a binomial distribution with parameters 6 and \(\frac{1}{2} .\) Show that \(X=3\) is the most likely outcome.

Short answer

The most likely outcome for a binomial distribution with parameters \(n = 6\) and \(p = \frac{1}{2}\) is \(X = 3\), as it has the highest probability value of \(\frac{20}{64}\).

Step-by-step solution

Compute binomial probabilities for x=0 to x=6

To find the most likely outcome, we need to compute the binomial probabilities for each value of x: For each value of x from 0 to 6, compute: \(P(X = x) = {6 \choose x} (\frac{1}{2})^x (1 - \frac{1}{2})^{6 - x}\)

Calculate PMF values for each possible outcome

We compute the following PMF values: \(P(X = 0) = {6 \choose 0} (\frac{1}{2})^0 (1 - \frac{1}{2})^{6 - 0} = 1 \cdot 1 \cdot (\frac{1}{2})^6 = \frac{1}{64}\) \(P(X = 1) = {6 \choose 1} (\frac{1}{2})^1 (1 - \frac{1}{2})^{6 - 1} = 6 \cdot \frac{1}{2} \cdot (\frac{1}{2})^5 = \frac{6}{64}\) \(P(X = 2) = {6 \choose 2} (\frac{1}{2})^2 (1 - \frac{1}{2})^{6 - 2} = 15 \cdot (\frac{1}{2})^2 \cdot (\frac{1}{2})^4 = \frac{15}{64}\) \(P(X = 3) = {6 \choose 3} (\frac{1}{2})^3 (1 - \frac{1}{2})^{6 - 3} = 20 \cdot (\frac{1}{2})^3 \cdot (\frac{1}{2})^3 = \frac{20}{64}\) \(P(X = 4) = {6 \choose 4} (\frac{1}{2})^4 (1 - \frac{1}{2})^{6 - 4} = 15 \cdot (\frac{1}{2})^4 \cdot (\frac{1}{2})^2 = \frac{15}{64}\) \(P(X = 5) = {6 \choose 5} (\frac{1}{2})^5 (1 - \frac{1}{2})^{6 - 5} = 6 \cdot (\frac{1}{2})^5 \cdot (\frac{1}{2})^1 = \frac{6}{64}\) \(P(X = 6) = {6 \choose 6} (\frac{1}{2})^6 (1 - \frac{1}{2})^{6 - 6} = 1 \cdot (\frac{1}{2})^6 \cdot 1 = \frac{1}{64}\)

Compare PMF values to find the most likely outcome

Comparing the computed PMF values: \(P(X = 0) = \frac{1}{64}\) \(P(X = 1) = \frac{6}{64}\) \(P(X = 2) = \frac{15}{64}\) \(P(X = 3) = \frac{20}{64}\) \(P(X = 4) = \frac{15}{64}\) \(P(X = 5) = \frac{6}{64}\) \(P(X = 6) = \frac{1}{64}\) The highest probability is for \(X = 3\), with a probability of \(\frac{20}{64}\). Hence, the most likely outcome is X = 3.