Question:medium

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?

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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!
Updated On: Jun 3, 2026
  • \(1/4 \)
  • \(1/3 \)
  • \(3/12 \)
  • \(5/12 \)
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Understanding the Concept:
Probability is a mathematical branch that deals with the likelihood of a specific event occurring. It is expressed as a numerical value between 0 and 1.
In this specific scenario, we are performing a single random draw. The "Total Outcomes" represent the entire collection of objects available in the bag.
The "Favorable Outcomes" are those that satisfy the given condition. The condition here is "neither red nor blue."
This logic implies that we must exclude the red balls and the blue balls, leaving only the remaining color category as our target.
Key Formula or Approach:
1. Probability \(P(E) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}\)
2. Total Outcomes = Sum of all items.
Step 2: Detailed Explanation:
First, we must establish the size of the total sample space. We do this by summing all the balls in the bag:
\[ \text{Total balls} = 5 \text{ (red)} + 4 \text{ (blue)} + 3 \text{ (green)} \]
\[ \text{Total Outcomes} = 12 \]
Second, we identify the favorable outcomes based on the constraint "neither red nor blue."
This means we subtract the red and blue balls from the total, or simply count the green balls:
\[ \text{Favorable Outcomes} = \text{Green Balls} = 3 \]
Now, applying the basic probability formula:
\[ P = \frac{3}{12} \]
In competitive examinations, fractions should always be reduced to their simplest form. We divide both numerator and denominator by their highest common factor, which is 3:
\[ \frac{3 \div 3}{12 \div 3} = \frac{1}{4} \]
Although 3/12 is mathematically equivalent, 1/4 is the standard simplified answer expected.
Step 3: Final Answer:
The probability of drawing a ball that is neither red nor blue is 1/4. Thus, Option (A) is the correct choice.
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