Colleen times her morning commute such that there is an equal likeliho

This problem asks you to calculate the probability that Colleen will arrive late to work no more than twice during a five-day workweek (so that is none, once, or twice). Since Colleen has an equal chance of being early or late (she is never on time), you essentially have a coin flip: equal chance between E and L on each day.

There are two ways to solve this problem. One of them is calculational, one of them is conceptual. Let's start with the calculational:

Method #1: Do the Math
Any probability question basically breaks down a single concept:
\(P=\frac{\text{# of situations that fit the criteria}}{\text{Total # of situations}}\)

Calculating the total number of possibilities is relatively straightforward using your knowledge that there are two possibilities each day. With \(5\) days having \(2\) options each, the total number of possible situations is \(2^5 = 32\).

Now you need the favorable options. First calculate the easy ones:

There is only one arrangement when she is early every day: (EEEEE).

There are five arrangements where she can be late once per week: (LEEEE, ELEEE, EELEE, EEELE, EEEEL).

The number of arrangements where Colleen is late twice requires a bit of combinatorics. In this case, you are dealing with a permutation with repeated elements (three Es and two Ls). Therefore, the number of arrangements is equal to:

\(A=\frac{N!}{R_E!∗R_L!}=\frac{5!}{3!∗2!}=\frac{5∗4∗3!}{3!∗2}=\frac{5∗4}{2}=10\)

Combining all of these options together, the total number of ways Colleen can arrange her week and still be late "no more than two times" is \(1 + 5 + 10 = 16\). To calculate the probability, we divide this number of the total number of possible arrangements (\(32\)):
\(P=\frac{\text{# of situations that fit the criteria}}{\text{Total # of situations}}=\frac{1+5+10}{32}=\frac{16}{32}=\frac{1}{2}\)

The answer is “C”.

Method #2: Conceptualize
If you can see it, there is a much quicker way to conceptually get to the answer. As we went through Method #1, did you notice the natural symmetry between being (1) always early and (2) always late, being (1) early once and (2) late once, and being (1) early twice and (1) late twice? Think about it: for every situation where Colleen is early a certain number of times, there is an equal number of options where she is late the same number of times. Since she can be late 0, 1, 2, 3, 4, or 5 times in a given week and "no more than two times" is half of those symmetrically-sized options, then she has a \(50\%\) chance (or \(3\) in \(6\)) chance of meeting the criteria given in the problem. The answer is still "C", but the question can be solved in less than 30 seconds.