If the probability of rain on any given day in City X is 50 percent

The probability of rain each day is 1/2 and the probability of no rain is also 1/2. \(C^3_5=10\) represent ways to choose on which 3 days out of 5 there will be a rain, so \(P=C^3_5*(\frac{1}{2})^3*(\frac{1}{2})^2=\frac{10}{32}=\frac{5}{16}\).

Answer: C.

Or think about it this way: we want the probability of the following event: RRRNN, where R represent rain day and N represents no-rain day. Now, each R and each N have individual probability of 1/2, so \((\frac{1}{2})^5\).

But the case of RRRNN can occur in many ways: RRRNN, RRNRRN, RNRRN, NRRRN, ... basically it will be equla to # of arrangements (permutations) of 5 letters RRRNN out of which there are 3 identical R's and 2 identical N's. That # of arrangements is \(\frac{5!}{3!2!}\), (notice that it's the same as \(C^3_5\)). So, finally \(P=\frac{5!}{3!2!}*(\frac{1}{2})^5=\frac{5}{16}\).

Answer: C.

For more on this topic check Combinations and Probability chapters of Math Book:
math-combinatorics-87345.html
math-probability-87244.html

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Hope it helps.