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?

A.

\(\frac{9}{20}\)

B.

\(\frac{3}{10}\)

C.

\(\frac{1}{4}\)

D.

\(\frac{11}{20}\)

Correct answer is A

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  \(\frac{9}{20}\)