Step 1: Listing the Sample Space Idea:
Since each of the 4 days independently has 2 equally likely outcomes, cloudy or sunny, the total number of equally likely outcome sequences over 4 days is $2^4 = 16$. We now need to count how many of these 16 equally likely sequences have exactly 3 cloudy days and 1 sunny day.
Step 2: Counting Favourable Outcomes Directly:
Instead of applying the binomial formula directly, we can list the position of the single sunny day among the four days, since the rest are automatically cloudy. The sunny day can be on day 1, day 2, day 3 or day 4, giving four distinct sequences, for example Sunny-Cloudy-Cloudy-Cloudy, Cloudy-Sunny-Cloudy-Cloudy, Cloudy-Cloudy-Sunny-Cloudy, and Cloudy-Cloudy-Cloudy-Sunny.
Step 3: Detailed Explanation:
Since there are exactly 4 favourable sequences out of the 16 total equally likely sequences of 4 independent cloudy or sunny outcomes, each individual sequence being equally likely because p = 0.5 for both cloudy and sunny, the probability is the ratio of favourable sequences to total sequences, which is $\frac{4}{16} = \frac{1}{4}$.
This matches exactly with the result obtained from the binomial coefficient method, $\binom{4}{3}(0.5)^3(0.5)^1 = 4 \times \frac{1}{16} = \frac{1}{4}$, confirming the answer through direct enumeration rather than only through the formula.
None of the other listed fractions, 3/4, 2/3 or 3/8, correspond to 4 favourable outcomes out of 16 total outcomes, so they can be ruled out by this direct counting method as well.
Final Answer:
By directly counting the 4 favourable sequences among the 16 equally likely possible sequences, the probability is confirmed to be 1/4.
$\boxed{\dfrac{1}{4}}$