Answer
a) $\frac{1}{10}$
b) $\frac{1}{2}$
c) $\frac{1}{2}$
Work Step by Step
Note that $S =$ {$95, 96, 97, 98, 99, 100, 101, 102, 103, 104$} and each possible outcome has probability $\frac{1}{10}$ of occuring.
a) We want to find $P(100) = \frac{1}{10}$
b) We want to find $P(less than 100) = P(95) + P(96) + P(97) + P(98) + P(99) = \frac{1}{10} + \frac{1}{10} + \frac{1}{10} + \frac{1}{10} + \frac{1}{10} = \frac{5}{10} = \frac{1}{2}$
c) We want to find $P(98\leq x\leq102) = P(98) + P(99) + P(100) + P(101) + P(102) = \frac{1}{10} + \frac{1}{10} + \frac{1}{10} + \frac{1}{10} + \frac{1}{10} = \frac{5}{10} = \frac{1}{2}$
