This problem can be modeled using a binomial distribution where the number of trials is n=4.
Let success be defined as a cloudy day:
Probability of a cloudy day, p=0.5=21
Probability of a sunny day, q=1−p=0.5=21
The probability of having exactly k=3 cloudy days (and therefore 1 sunny day) in n=4 independent days is:
P(X=3)=(kn)pkqn−k=(34)(21)3(21)1 Evaluating this:
P(X=3)=4×81×21=4×161=41 Hence, the correct option is (A).