This problem involves calculating probability for a sequence of independent events, which can be solved using the binomial probability formula.

Binomial Probability Calculation

We are given:

• The probability of a day being cloudy, \( p = 0.5 \).
• The probability of a day being sunny, \( q = 1 - p = 1 - 0.5 = 0.5 \).
• The number of days observed, \( n = 4 \).
• We want to find the probability of exactly 3 cloudy days (successes) and 1 sunny day (failure). Let \( k = 3 \).

The binomial probability formula is:

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

Here, \( \binom{n}{k} \) represents the number of combinations of choosing \( k \) successes from \( n \) trials.

Applying the Formula

1. Calculate Combinations: Find the number of ways to choose 3 cloudy days out of 4. $ \binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4!}{3!1!} = \frac{4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(1)} = 4 $
2. Calculate Probability of Specific Sequence: For any specific sequence with 3 cloudy days and 1 sunny day (e.g., CCCS), the probability is: $ p^3 q^1 = (0.5)^3 (0.5)^1 = (0.5)^4 $
3. Calculate Total Probability: Multiply the number of combinations by the probability of one specific sequence. $ P(3 \text{ cloudy days}) = \binom{4}{3} \times (0.5)^3 \times (0.5)^1 $ $ P(3 \text{ cloudy days}) = 4 \times (0.5)^4 $ $ P(3 \text{ cloudy days}) = 4 \times \left(\frac{1}{2}\right)^4 $ $ P(3 \text{ cloudy days}) = 4 \times \frac{1}{16} $ $ P(3 \text{ cloudy days}) = \frac{4}{16} = \frac{1}{4} $

Final Probability

The probability of having exactly three cloudy days and one sunny day in any given four days is 1/4.