A simple classical probability theory misunderstanding

Solution 1:

You can look at this in two ways:

1. The sample space is $ \{ M, I, S, S, I, S, S, I, P, P, I \} $ and the simple events of the sample space all have equal probabilities. Using $ P(A) = \dfrac{n_A}{n_S} $ we have $ P(A) = \dfrac{2}{11} $.

2. The sample space is $ \{ M , I , S, P \} $ but the simple events of the sample space do not have equal probabilities. This means that we must asign probabilities to all four $ \{M\}, \{I\}, \{S\} , \{P\} $. We can actually assign any numbers we want provided they are non-negative and add up to 1. Of course the only assignment that corresponds to the problem you are describing is $ P(\{M\}) = \frac{1}{11}, ~P(\{I\})=\frac{4}{11}, ~P(\{S\})=\frac{4}{11} , P(\{P\})=\frac{2}{11} $.