Question 1099537
In this problem, you must make sure to avoid double counting. In the first question, when we compute the probability that a ball is red, we have already counted both the shiny red balls and the dull red balls.  Thus, when we compute the probability of getting a shiny ball, we must only consider the ones that are blue since the shiny red ones were already counted.
Thus P(red or shiny) = P(red) + P(shiny blue)
P(red) = 79/200
The number of shiny blue balls is 91 - 55 = 36
P(shiny blue) = 36/200
So P(red or shiny) = (79+36)/200 = 0.575
The number of dull blue balls = 200 - 79 - 36 = 85
Therefore P(dull blue) = 85/200 = 0.425
Notice that the two probabilities add to 1, as they should since a ball is either red, shiny blue, or dull blue