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In the world of probability, a probability function is key to predicting outcomes. For instances like this exercise, where we're dealing with a Binomial distribution, the probability function allows us to determine the likelihood of each possible result.

The specific formula for a binomial probability is:

• \( \Pr(Y = k) = \binom{n}{k} p^k (1-p)^{n-k} \)

Here, \(n\) is the number of trials, \(k\) is the number of successful trials we're interested in, and \(p\) is the probability of a single success.

In our example, with four judges (trials) and a one-third chance that any given judge will select formula \(B\), the formula helps us find the probability for \(Y\), the number of judges choosing formula \(B\). Each probability, for \(Y = 0, 1, 2, 3,\) and \(4\), describes how likely it is that exactly that many judges preferred the new formula.

The expected value in probability gives us the average outcome we'd expect if we repeated the experiment many times. It's a way to summarize possible outcomes into a single, comprehensible figure.

For a binomial distribution, the expected value \(E[Y]\) is simply \(np\). In our scenario, this translates to:

• The number of trials \( (n) = 4\)
• The probability of one judge selecting formula \(B\) \((p = \frac{1}{3})\)

With these, we calculate:

• \( E[Y] = 4 \times \frac{1}{3} = \frac{4}{3} \approx 1.33\)

This expected value tells us that if the experiment with the judges were conducted many times, on average about 1.33 judges would choose the new formula \(B\).

It's crucial to remember that the expected value may not always be a possible actual result (like having a fraction of a judge in this case), but it's invaluable for understanding trends over many trials.

Variance provides a measure of how much the results of a probability experiment are spread out. It tells us how much the counts of the judges believing formula \(B\) might deviate from our expected value, or average count.

For the binomial distribution, variance is computed using the formula:

• \( \text{Var}(Y) = np(1-p) \)

In this instance, \(n = 4\) and \(p = \frac{1}{3}\), which gives us:

• \( \text{Var}(Y) = 4 \times \frac{1}{3} \times \frac{2}{3} = \frac{8}{9} \approx 0.89\)

This variance is a measure of how varied the outcomes of the judges’ choices are. A higher variance would indicate that judges' preferences are more spread out, while a lower variance points to closer clustering around the expected value.

Understanding variance is crucial for predicting the consistency of results and for assessing the reliability of your expected value.