A box contains 40 identical balls. Among these, 10 are red and 12 are blue. A ball is randomly selected from the box. What is the probability that the selected ball is neither red nor blue?

1. Problem Description

A box contains 40 identical balls. Among these, 10 are red and 12 are blue. A ball is randomly selected from the box. What is the probability that the selected ball is neither red nor blue?

2. Solution Steps

The total number of balls in the box is

4

0. The number of red balls is

1

0. The number of blue balls is

1

2. The number of balls that are either red or blue is $10 + 12 = 22$.

The number of balls that are neither red nor blue is $40 - 22 = 18$.

The probability of selecting a ball that is neither red nor blue is the number of balls that are neither red nor blue divided by the total number of balls:

$P(\text{neither red nor blue}) = \frac{\text{Number of balls that are neither red nor blue}}{\text{Total number of balls}} = \frac{18}{40}$

We can simplify this fraction by dividing both numerator and denominator by their greatest common divisor, which is 2:

$\frac{18}{40} = \frac{18 \div 2}{40 \div 2} = \frac{9}{20}$

3. Final Answer

The probability that the selected ball is neither red nor blue is $\frac{9}{20}$.

So the answer is A. $\frac{9}{20}$