91Ó°ÊÓ

First, let's define the given parameters for the problem. We know that \(X\) has a binomial distribution with \(n = 6\) trials and success probability \(p = 1/2\).

Now we need to calculate the probability of each possible outcome for \(X\), from 0 to 6. We can do this using the given probability mass function. For \(k = 0\): \[P(X=0) = {6 \choose 0} \left(\frac{1}{2}\right)^0 \left(1-\frac{1}{2}\right)^{6-0}\] For \(k = 1\): \[P(X=1) = {6 \choose 1} \left(\frac{1}{2}\right)^1 \left(1-\frac{1}{2}\right)^{6-1}\] For \(k = 2\): \[P(X=2) = {6 \choose 2} \left(\frac{1}{2}\right)^2 \left(1-\frac{1}{2}\right)^{6-2}\] For \(k = 3\): \[P(X=3) = {6 \choose 3} \left(\frac{1}{2}\right)^3 \left(1-\frac{1}{2}\right)^{6-3}\] For \(k = 4\): \[P(X=4) = {6 \choose 4} \left(\frac{1}{2}\right)^4 \left(1-\frac{1}{2}\right)^{6-4}\] For \(k = 5\): \[P(X=5) = {6 \choose 5} \left(\frac{1}{2}\right)^5 \left(1-\frac{1}{2}\right)^{6-5}\] For \(k = 6\): \[P(X=6) = {6 \choose 6} \left(\frac{1}{2}\right)^6 \left(1-\frac{1}{2}\right)^{6-6}\]

Now that we have calculated the probabilities of each possible outcome, let's compare them to see which one is the most likely: \[P(X=0) = {6 \choose 0} \left(\frac{1}{2}\right)^0 \left(1-\frac{1}{2}\right)^{6-0} = 0.015625\] \[P(X=1) = {6 \choose 1} \left(\frac{1}{2}\right)^1 \left(1-\frac{1}{2}\right)^{6-1} = 0.09375\] \[P(X=2) = {6 \choose 2} \left(\frac{1}{2}\right)^2 \left(1-\frac{1}{2}\right)^{6-2} = 0.234375\] \[P(X=3) = {6 \choose 3} \left(\frac{1}{2}\right)^3 \left(1-\frac{1}{2}\right)^{6-3} = 0.3125\] \[P(X=4) = {6 \choose 4} \left(\frac{1}{2}\right)^4 \left(1-\frac{1}{2}\right)^{6-4} = 0.234375\] \[P(X=5) = {6 \choose 5} \left(\frac{1}{2}\right)^5 \left(1-\frac{1}{2}\right)^{6-5} = 0.09375\] \[P(X=6) = {6 \choose 6} \left(\frac{1}{2}\right)^6 \left(1-\frac{1}{2}\right)^{6-6} = 0.015625\] The highest probability is for \(X = 3\): \(P(X=3) = 0.3125\) Since the probability for \(X = 3\) is the highest of all possible outcomes, we can conclude that \(X = 3\) is the most likely outcome.