A box contains 40 identical balls of which 10 are red and 12 are blue. if a ball is selected at random from the box what is the probability that it is neither red nor blue?
Total number = 40
number of red = 10
Pr (R) = \(\frac{10}{40}\)
Pr(blue) = \(\frac{12}{40}\)
probability (neither red nor blue)
= 1 - \(\frac{10}{40}\) - \(\frac{12}{40}\)
= \(\frac{40 -12 -10}{40}\)
= \(\frac{18}{40}\) or \(\frac{9}{20}\)
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