A box contains 40 identical balls of which 10 are red and 12 are blue. if a ball is selected at random from…

Analyze the question: We need the probability that a ball drawn at random is neither red nor blue, so it must be some other colour.

Concept: Probability = (number of favourable outcomes) ÷ (total number of outcomes).

Solution:

• Total balls = 40
• Red = 10, Blue = 12
• Red or blue = 10 + 12 = 22
• Neither red nor blue = 40 − 22 = 18

P(neither red nor blue) = 18/40 = 9/20

Check: P(red) = 10/40 = 1/4 and P(blue) = 12/40 = 3/10. Together: 1/4 + 3/10 = 5/20 + 6/20 = 11/20. Then 1 − 11/20 = 9/20. ✔

Answer: A. 9/20

Note that D (11/20) is the probability of getting a red or blue ball. It is the trap answer for students who forget to subtract from 1.

Exam tip: For "neither... nor" questions, first count the outcomes you do not want, then subtract them from the total (or use 1 − P(unwanted)). Always simplify your fraction to match the options.