If the lady cannot differentiate the taste, her guesses are purely random, and the problem boils down to calculating the probability of making exactly $k = 12$ correct guesses out of $n = 16$ total cups.

This is a binomial probability problem, where:

• $n = 16$: The total number of cups.
• $p = 0.5$: The probability of guessing correctly for each cup.
• $k = 12$: The desired number of correct guesses.

The binomial probability formula is given by:

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

Where:

• $\binom{n}{k} = \frac{n!}{k!(n-k)!}$: The number of ways to choose $k$ successes from $n$ trials.
• $p = 0.5$: The probability of success in each trial.

Step 1: Compute the binomial coefficient

$\binom{16}{12} = \frac{16!}{12!(16-12)!} = \frac{16 \times 15 \times 14 \times 13}{4 \times 3 \times 2 \times 1} = 1820$

Step 2: Compute the probability

Substitute $n = 16$, $k = 12$, and $p = 0.5$ into the formula:

$P(X = 12) = \binom{16}{12} (0.5)^{12} (0.5)^{16-12}$

$P(X = 12) = 1820 \cdot (0.5)^{12} \cdot (0.5)^4$

$P(X = 12) = 1820 \cdot (0.5)^{16}$

$P(X = 12) = 1820 \cdot \frac{1}{65536}$

$P(X = 12) = \frac{1820}{65536} \approx 0.0278$

Final Answer:

The probability of making exactly 12 correct guesses is approximately 0.0278 (rounded to 4 decimal places).

Let me know if you'd like further clarification or additional calculations! 😊

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Here are 5 related questions for further exploration:

1. What is the probability of making exactly 8 correct guesses?
2. How does the probability change if $p \neq 0.5$, assuming bias in the guesses?
3. What is the expected number of correct guesses in this scenario?
4. What is the probability of making at least 12 correct guesses?
5. How would the probability change if there were 20 cups instead of 16?

Tip: For large values of $n$, approximations like the Normal or Poisson distribution can simplify binomial probability calculations!