A box contains dark and white chocolate candies. Each type of chocolate candies can be either round or square. There is an $\frac{8}{15}$ probability of selecting a dark chocolate candy, a $\frac{9}{20}$ probability of selecting a round chocolate candy, and a *$\frac{1}{5}$** *probability of selecting a round and dark chocolate candy. What is the probability that a randomly selected chocolate candy is neither dark nor round?

$\frac{47}{60}$

$\frac{8}{15}$

$\frac{9}{20}$

$\frac{13}{60}$

$\frac{7}{60}$

**Solution:**

Let’s define A as *chocolate is dark* and B as *chocolate is round*. Using the above formula, we have:

$\Rightarrow $ P(A) = $\frac{8}{15}$

$\Rightarrow $ P(B) = $\frac{9}{20}$

We are told that the probability of selecting a *round and dark chocolate* candy is $\frac{1}{5}$:

$\Rightarrow $ P(both A and B) = $\frac{1}{5}$

$\Rightarrow $ 1 = P(A) + P(B) - P(both A and B) + P(neither A nor B)

$\Rightarrow $ 1 = $\frac{8}{15}+\frac{9}{20}-\frac{1}{5}+P\left(\text{neither A nor B}\right)$

$\Rightarrow $ 1 = $\frac{32}{60}+\frac{27}{60}-\frac{12}{60}+P\left(\text{neitherAnorB}\right)$

$\Rightarrow $ 1 = $\frac{47}{60}+P\left(\text{neither A nor B}\right)$

Therefore, the probability that a randomly selected chocolate is neither dark nor round is $1-\frac{47}{60}=\frac{13}{60}$.