Question

A bag contains 5 red, 4 blue, and 3 green balls. If one ball is drawn at random, what is the probability that it is neither red nor blue?

• A. \(1/4 \)
• B. \(1/3 \)
• C. \(3/12 \)
• D. \(5/12 \)

Hint:

To tackle "neither/nor" probability statements rapidly, subtract the combined probability of the unwanted items from the whole number 1: - \(P(\text{Neither A nor B}) = 1 - [P(A) + P(B)]\) - \(\text{Probability of Red or Blue} = \frac{5 + 4}{12} = \frac{9}{12}\) - \(\text{Probability of Neither} = 1 - \frac{9}{12} = \frac{3}{12} = \frac{1}{4}\) This complementary technique prevents calculation errors when working with large quantities of items!

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Probability represents the numerical likelihood of a specific outcome taking place, calculated as the ratio of favorable outcomes to the total pool of possible outcomes. Here, the condition "neither red nor blue" means that the selected ball must belong to an alternative color family. In this bag, that leaves only the green balls as our valid favorable outcomes.

Step 2: Key Formula or Approach:
\[ \text{Probability of an Event } P(E) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}} \]

Step 3: Detailed Explanation:
First, find the total number of balls present inside the bag to establish the total number of possible outcomes: \[ \text{Total Outcomes} = 5 \text{ (Red)} + 4 \text{ (Blue)} + 3 \text{ (Green)} = 12 \text{ balls} \] Next, determine the number of favorable outcomes. Because the ball cannot be red and cannot be blue, it must be green: \[ \text{Favorable Outcomes (Green Balls)} = 3 \] Now, substitute these parameters into our main probability equation: \[ P(\text{Neither Red nor Blue}) = \frac{3}{12} \] To simplify the fraction, divide both the numerator and the denominator by their greatest common divisor, which is \(3\): \[ \frac{3 \div 3}{12 \div 3} = \frac{1}{4} \] While both \(\frac{1}{4}\) (Option A) and \(\frac{3}{12}\) (Option C) represent the exact same mathematical value, standard competitive exams typically favor the lowest reduced fraction, making Option (A) the formal correct answer choice.

Step 4: Final Answer:
The probability that the drawn ball is neither red nor blue is 1/4.