Question

A Bag contains 5 white balls and 7 red balls. A ball is drawn at random from the bag. The probability that it is either a white ball or red ball is :

• A. \(\frac{5}{12}\)
• B. \(\frac{7}{12}\)
• C. 1
• D. \(\frac{5}{7}\)

Hint:

The bag only contains white and red balls. So, if you draw one ball, it is {certain} to be either white or red. The probability of a certain event is always 1. Total balls = 5 (white) + 7 (red) = 12. Favorable outcomes (ball is white or red) = 12. Probability = \(\frac{12}{12} = 1\).

Answer 1

The Correct Option is C

Concept: Probability of an event = \(\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}\). If an event is certain to happen, its probability is 1. Step 1: Determine the total number of balls in the bag Number of white balls = 5 Number of red balls = 7 Total number of balls = Number of white balls + Number of red balls Total number of balls = \(5 + 7 = 12\). This is the total number of possible outcomes when one ball is drawn. Step 2: Identify the event for which probability is to be found The event is "the ball drawn is either a white ball or a red ball." Step 3: Determine the number of favorable outcomes Since the bag only contains white balls and red balls, any ball drawn from the bag will {always} be either white or red. Number of white balls = 5 Number of red balls = 7 Number of balls that are "either white or red" = \(5 + 7 = 12\). So, the number of favorable outcomes is 12. Step 4: Calculate the probability Probability (either white or red) = \(\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}\) \[ P(\text{either white or red}) = \frac{12}{12} \] \[ P(\text{either white or red}) = 1 \] This means the event is a certain event. If you draw a ball, it must be one of the colors present in the bag. This matches option (3). {Alternative approach using probability rules:} Let W be the event of drawing a white ball, and R be the event of drawing a red ball. \(P(W) = \frac{5}{12}\) \(P(R) = \frac{7}{12}\) The events W and R are mutually exclusive (a ball cannot be both white and red at the same time). The probability of "either W or R" is \(P(W \cup R) = P(W) + P(R)\) for mutually exclusive events. \(P(W \cup R) = \frac{5}{12} + \frac{7}{12} = \frac{5+7}{12} = \frac{12}{12} = 1\).